Created by Team Vardaan | VARDAAN COMET | Reference: ML Agarwal – ICSE Class 9

Triangles

Syllabus at a Glance

ICSE Class 9 – Chapter 9: Triangles

Part A – Congruency

Four congruence rules: SSS, SAS, AAS, RHS. Illustration through cutouts. Simple applications.

Part B – Properties and Inequalities

Angles opposite equal sides; sides opposite greater angles; triangle inequality; shortest distance from external point.

Note: Proofs are not required for ICSE examination. Focus on correct identification and application of results.

1. Revision of Basic Concepts

What is a Triangle?

[Diagram Placeholder]
A clean precise mathematical diagram. Background color: #e8f5e9 (light green, to match this box). Triangle ABC: vertex A at top, B at bottom-left, C at bottom-right, all on a #e8f5e9 background. Sides labelled: a = BC (opposite A), b = CA (opposite B), c = AB (opposite C). Interior angles marked with dark green arc symbols labelled angle-A, angle-B, angle-C. One exterior angle at B drawn by extending AB beyond B, marked in red, labelled "Exterior angle = angle-A + angle-C". Thin dark lines, bold sans-serif font. Background strictly #e8f5e9.

A triangle is a closed plane figure formed by three line segments joining three non-collinear points.

Triangle $ABC$ has vertices $A$, $B$, $C$; sides $AB$, $BC$, $CA$; and angles $\angle A$, $\angle B$, $\angle C$.
Side opposite vertex $A$: $a = BC$; opposite $B$: $b = CA$; opposite $C$: $c = AB$.
Angle Sum Property: $\angle A + \angle B + \angle C = 180°$.
Exterior Angle Property: An exterior angle of a triangle equals the sum of the two non-adjacent interior angles.

Types of Triangles

Classification	Type	Property
By Sides	Equilateral	All three sides equal; each angle = 60°
	Isosceles	Two sides equal; base angles equal
	Scalene	All three sides and angles different
By Angles	Acute-angled	All angles less than 90°
	Right-angled	One angle = 90°; other two complementary
	Obtuse-angled	One angle greater than 90°

2. Congruence of Triangles

Core Definition

[Diagram Placeholder]
A precise mathematical diagram. Background color: #e3f2fd (light blue, to match this box). Two congruent triangles side by side on a #e3f2fd background. Left: triangle ABC (A top, B bottom-left, C bottom-right). Right: triangle DEF (D top, E bottom-left, F bottom-right). Large congruence symbol ≅ between them. Equal sides: AB = DE (one red tick), BC = EF (two blue ticks), CA = FD (three green ticks). Equal angles: angle-A = angle-D (one blue arc), angle-B = angle-E (two red arcs), angle-C = angle-F (three green arcs). Clean black outlines. Background strictly #e3f2fd.

Two triangles are congruent if they are exactly the same in shape and size. One can be placed exactly over the other so that all corresponding sides and angles match perfectly.

Symbol: $\cong$
$\triangle ABC \cong \triangle DEF$ means: $AB = DE$, $BC = EF$, $CA = FD$, $\angle A = \angle D$, $\angle B = \angle E$, $\angle C = \angle F$.
CPCT (Corresponding Parts of Congruent Triangles) are equal — used as a reason after proving congruence.

The order of vertices matters. $\triangle ABC \cong \triangle DEF$ means $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$ strictly.

Illustration Through Cutouts

Practical Activity

Cut out two triangles from cardboard with identical measurements. Try placing one over the other by rotating, sliding, or flipping. If they match exactly (cover each other completely), the triangles are congruent. Flipping also works — this shows that congruent triangles can be mirror images of each other, which is key to understanding the RHS case.

3. The Four Cases of Congruency

There are four standard criteria to prove two triangles congruent.

Case 1 – SSS (Side – Side – Side)

[Diagram Placeholder]
A clear mathematical diagram. Background color: #e3f2fd (light blue, to match this box). Two identical scalene triangles side by side on a #e3f2fd background: triangle ABC left, triangle DEF right. Equal sides: AB = DE (one red tick each), BC = EF (two blue ticks each), CA = FD (three green ticks each). Bold label "SSS Congruence Rule" below. Background strictly #e3f2fd.

Statement: If three sides of one triangle are equal to the corresponding three sides of another triangle, the triangles are congruent.

Condition: $AB = DE$, $BC = EF$, $CA = FD$.

Conclusion: $\triangle ABC \cong \triangle DEF$ (by SSS)

If all three sides are equal, the triangle is completely rigid — no other triangle with those three lengths is possible.

Worked Example — SSS

Q. In $\triangle PQR$ and $\triangle XYZ$: $PQ = XY = 5$ cm, $QR = YZ = 6$ cm, $PR = XZ = 4$ cm. Prove $\triangle PQR \cong \triangle XYZ$ and identify $\angle P$'s equal in $\triangle XYZ$.

In $\triangle PQR$ and $\triangle XYZ$:
$PQ = XY = 5$ cm (given) …(i)
$QR = YZ = 6$ cm (given) …(ii)
$PR = XZ = 4$ cm (given) …(iii)

By SSS: $\triangle PQR \cong \triangle XYZ$.

By CPCT: $P \leftrightarrow X$ so $\angle P = \angle X$. Hence $\angle X$ equals $\angle P$.

Case 2 – SAS (Side – Angle – Side)

[Diagram Placeholder]
A precise mathematical diagram. Background color: #e3f2fd (light blue, to match this box). Two congruent triangles on a #e3f2fd background: ABC left, DEF right. Equal sides: AB = DE (one red tick), BC = EF (two blue ticks). Included angle at B and E marked with a green arc labelled "included angle". Label "SAS Congruence Rule" bold below. Background strictly #e3f2fd.

Statement: If two sides and the included angle of one triangle are equal to those of another, the triangles are congruent.

Condition: $AB = DE$, $\angle B = \angle E$ (included angle between AB, BC and DE, EF), $BC = EF$.

Conclusion: $\triangle ABC \cong \triangle DEF$ (by SAS)

The angle MUST be the included angle — the one between the two given sides. SSA (non-included angle) is NOT valid.

Worked Example — SAS

Q. In $\triangle ABD$ and $\triangle ACD$: $AB = AC$, AD bisects $\angle BAC$, $AD$ is common. Prove $\triangle ABD \cong \triangle ACD$ and hence show $AD \perp BC$.

In $\triangle ABD$ and $\triangle ACD$:
$AB = AC$ (given) …(i)
$\angle BAD = \angle CAD$ (AD bisects $\angle A$) …(ii)
$AD = AD$ (common) …(iii)

By SAS: $\triangle ABD \cong \triangle ACD$.

By CPCT: $BD = CD$ and $\angle ADB = \angle ADC$. Since $\angle ADB + \angle ADC = 180°$: $\angle ADB = 90°$, so $AD \perp BC$.

Case 3 – AAS (Angle – Angle – Side)

[Diagram Placeholder]
A precise mathematical diagram. Background color: #e3f2fd (light blue, to match this box). Two congruent scalene triangles on a #e3f2fd background: ABC left, DEF right. Equal angles: angle-A = angle-D (one blue arc each), angle-B = angle-E (two red arcs each). One equal non-included side: BC = EF (two green ticks) labelled "non-included side". Write "AAS Congruence Rule" bold below. Background strictly #e3f2fd.

Statement: If two angles and any one corresponding side (not necessarily the included side) of one triangle are equal to those of another, the triangles are congruent.

Condition: $\angle A = \angle D$, $\angle B = \angle E$, and one side (e.g., $BC = EF$ or $AB = DE$).

Conclusion: $\triangle ABC \cong \triangle DEF$ (by AAS)

If two angles are known, the third is fixed (angle sum = 180°). AAS completely determines the triangle once any one side is given.

Worked Example — AAS

Q. In $\triangle ABC$ and $\triangle DEF$: $\angle A = \angle D = 50°$, $\angle B = \angle E = 70°$, $BC = EF = 8$ cm. Prove $\triangle ABC \cong \triangle DEF$.

$\angle C = 180° - 50° - 70° = 60° = \angle F$.

In $\triangle ABC$ and $\triangle DEF$:
$\angle A = \angle D = 50°$ …(i)
$\angle B = \angle E = 70°$ …(ii)
$BC = EF = 8$ cm …(iii)

By AAS: $\triangle ABC \cong \triangle DEF$. By CPCT: $AB = DE$, $AC = DF$.

Case 4 – RHS (Right angle – Hypotenuse – Side)

[Diagram Placeholder]
A precise mathematical diagram. Background color: #e3f2fd (light blue, to match this box). Two congruent right triangles on a #e3f2fd background: ABC with right angle at B (small square at B), DEF with right angle at E (small square at E). Hypotenuse AC = DF (one red tick each, labelled "Hypotenuse"). Side BC = EF (two blue ticks). Label "RHS Congruence Rule" bold below. Background strictly #e3f2fd.

Statement: In two right-angled triangles, if the hypotenuse and one other side of one triangle are equal to those of the other, the triangles are congruent.

Applies ONLY to: Right-angled triangles.

If $\angle B = \angle E = 90°$, $AC = DF$ (hypotenuses), $BC = EF$, then $\triangle ABC \cong \triangle DEF$ (by RHS).

Hypotenuse = side opposite the right angle = longest side in a right triangle. Always identify it before applying RHS.

Worked Example — RHS

Q. $\angle ABC = \angle EDC = 90°$, $AC = EC$ (hypotenuses), $BC = DC$. Prove $\triangle ABC \cong \triangle EDC$ and show $AB = ED$.

In $\triangle ABC$ and $\triangle EDC$:
$\angle ABC = \angle EDC = 90°$ (given) …(i)
$AC = EC$ (hypotenuses equal) …(ii)
$BC = DC$ (sides equal) …(iii)

By RHS: $\triangle ABC \cong \triangle EDC$.

By CPCT: $AB = ED$.

Quick Comparison of the Four Cases

Rule	What Must Be Equal?	Key Condition	Applies To
SSS	All 3 sides	No angle condition needed	All triangles
SAS	2 sides + included angle	Angle must be between the two sides	All triangles
AAS	2 angles + any 1 side	Side need not be included	All triangles
RHS	Right angle + hypotenuse + 1 side	Must be right-angled triangles	Right triangles only

What Does NOT Work?

AAA: Proves similarity only, NOT congruence. Equal angles can have different sizes.
SSA: Non-included angle — gives ambiguous case (two triangles possible). Exception: RHS is the valid special case.

✏ Practice Problems — Congruence of Triangles

In $\triangle ABC$ and $\triangle PQR$: $AB = PQ = 7$ cm, $BC = QR = 5$ cm, $\angle B = \angle Q = 60°$. Name the congruence rule and write the congruence relation.
$O$ is the midpoint of both $AB$ and $CD$. Prove $\triangle AOC \cong \triangle BOD$ and state what follows about $AC$ and $BD$.
$\triangle ABC$ is isosceles with $AB = AC$. $D$ is the midpoint of $BC$. Prove $\triangle ABD \cong \triangle ACD$ (SSS). Hence show $AD \perp BC$.
Right triangles $ABC$ and $DEF$: right angles at $C$ and $F$, hypotenuse $AB = DE$, leg $BC = EF$. Prove $\triangle ABC \cong \triangle DEF$ (RHS). Hence show $\angle A = \angle D$.
$\triangle PQR$ and $\triangle XYZ$: $\angle P = \angle X = 40°$, $\angle R = \angle Z = 80°$, $QR = YZ = 6$ cm. Prove congruence and identify the rule used.
$ABCD$ is a square. Prove diagonal $AC$ divides it into two congruent triangles. State the rule.
$AB \parallel CD$, $AB = CD$, $M$ is midpoint of $BC$. Prove $\triangle ABM \cong \triangle DCM$.
In $\triangle ABC$, $D$ on $BC$ with $AD$ bisecting $\angle BAC$ and $AB = AC$. Prove $\angle ADB = \angle ADC = 90°$.
$P$ is equidistant from two lines $l$ and $m$ intersecting at $A$. Prove $AP$ bisects the angle between $l$ and $m$.
$PS \perp QR$ and $QS = SR$. Prove $PQ = PR$. What type of triangle is $\triangle PQR$?

4. Angles Opposite Equal Sides Are Equal (and Converse)

Theorem 1 — Isosceles Triangle Theorem

[Diagram Placeholder]
A clean mathematical diagram. Background color: #fff3e0 (light orange, to match this theorem box). Isosceles triangle ABC on a #fff3e0 background: A at top, B at bottom-left, C at bottom-right. Equal sides AB and AC with single red tick marks. Equal base angles at B and C with double dark-orange arc symbols. Labels "AB = AC" pointing to equal sides, "angle-B = angle-C" pointing to base angles. Below: "Angles opposite equal sides are equal." Bold sans-serif. Background strictly #fff3e0.

Statement: If two sides of a triangle are equal, the angles opposite those sides are also equal.

In $\triangle ABC$: $AB = AC \Rightarrow \angle B = \angle C$.

Proof not required for ICSE examination.

Converse — Equal Angles Imply Equal Opposite Sides

Statement: If two angles of a triangle are equal, the sides opposite those angles are equal.

In $\triangle ABC$: $\angle B = \angle C \Rightarrow AB = AC$ (triangle is isosceles).

Important Corollaries

Equilateral triangle: All sides equal $\Rightarrow$ all angles = 60° (and converse).
Isosceles formula: With $AB = AC$ and apex $\angle A = x°$, base angles $= \dfrac{180° - x°}{2}$.

Worked Example 1

Q. In $\triangle ABC$, $AB = AC$, $\angle BAC = 50°$. Find $\angle ABC$ and $\angle ACB$.

Let $\angle ABC = \angle ACB = x$ (angles opp. equal sides).

$50° + 2x = 180°$ ⇒ $x = 65°$.

Answer: $\angle ABC = \angle ACB = 65°$.

Worked Example 2

Q. In $\triangle PQR$, $\angle Q = \angle R = 55°$. Find $\angle P$ and state which sides are equal.

$\angle P = 180° - 55° - 55° = 70°$.

$\angle Q = \angle R$ ⇒ sides opposite them equal: $PR = PQ$ (converse).

Answer: $\angle P = 70°$, $PQ = PR$.

Worked Example 3 (ML Agarwal Type)

Q. $\triangle ABC$ is isosceles with $AB = AC$. $D$, $E$ on $BC$ with $BD = CE$. Prove $AD = AE$.

$AB = AC$ ⇒ $\angle ABD = \angle ACE$ (base angles) …(i)

In $\triangle ABD$ and $\triangle ACE$: $AB = AC$ …(ii); $\angle ABD = \angle ACE$ …(iii); $BD = CE$ …(iv).

By SAS: $\triangle ABD \cong \triangle ACE$. By CPCT: $AD = AE$. ∴ proved.

Worked Example 4

Q. $\triangle ABC$: $AB = AC$. $P$ on $AB$, $Q$ on $AC$ with $AP = AQ$. Prove $BQ = CP$.

$BP = AB - AP = AC - AQ = CQ$ (since $AB = AC$, $AP = AQ$) …(i)

$AB = AC$ ⇒ $\angle ABQ = \angle ACB$ …(ii) [same as base angles]

In $\triangle ABQ$ and $\triangle ACP$: $AB = AC$; $\angle B = \angle C$; $AQ = AP$.
By SAS: $\triangle ABQ \cong \triangle ACP$. By CPCT: $BQ = CP$.

✏ Practice Problems — Isosceles Triangle and Equal Angles

In $\triangle ABC$, $AB = BC$, $\angle BAC = 40°$. Find all angles.
Isosceles $\triangle PQR$ with $PQ = PR$, vertex angle $\angle P = 80°$. Find the base angles.
In $\triangle ABC$, $\angle B = \angle C = 65°$. Which sides are equal? Justify.
In $\triangle ABC$, $AB = AC$. The bisector of $\angle B$ meets $AC$ at $D$. Prove $BD = BC$.
In $\triangle PQR$, $PQ > PR$. $S$ is on $QR$ with $PS$ bisecting $\angle QPR$. Show $\angle PSQ > \angle PSR$.
If the bisector of the vertical angle of a triangle bisects the base, prove the triangle is isosceles.
$ABCD$ is a square, $\triangle APB$ is equilateral ($P$ inside). Prove $\triangle APD \cong \triangle BPC$ and $DP = CP$.
In $\triangle ABC$, $\angle B = \angle C$. $D$ on $AB$, $E$ on $AC$ with $BD = CE$. Prove $\triangle DBC \cong \triangle ECB$ and $DC = EB$.
Sides $AB$, $BC$ and median $BM$ to $AC$ of $\triangle ABC$ equal sides $PQ$, $QR$ and median $QN$ to $PR$ of $\triangle PQR$. Prove $\triangle ABC \cong \triangle PQR$.
Isosceles $\triangle ABC$, $AB = AC$. Altitudes $BD$ to $AC$ and $CE$ to $AB$. Prove $BD = CE$.

5. Greater Angle is Opposite Greater Side (and Converse)

Theorem 2 — Side-Angle Inequality

[Diagram Placeholder]
A clean mathematical diagram. Background color: #fff3e0 (light orange, to match this theorem box). Scalene triangle ABC on a #fff3e0 background: A at top-left, B at bottom-left, C at bottom-right. Sides labelled: BC = 9 cm (longest, opposite A), AB = 7 cm (opposite C), AC = 5 cm (shortest, opposite B). Largest angle at A (big red arc, labelled "angle-A largest"), medium angle at C (blue arc), smallest angle at B (green arc). Two-headed arrows: "angle-A (largest) ↔ BC (longest)" and "angle-B (smallest) ↔ AC (shortest)". Background strictly #fff3e0.

Statement: If two sides of a triangle are unequal, the angle opposite the longer side is greater.

In $\triangle ABC$: $AB > AC \Rightarrow \angle C > \angle B$ (angle opposite $AB$ is greater).

Proof not required for ICSE examination.

Converse — Greater Angle Opposite Greater Side

Statement: If two angles of a triangle are unequal, the side opposite the greater angle is longer.

In $\triangle ABC$: $\angle A > \angle B \Rightarrow BC > AC$.

Worked Example 1

Q. $\triangle ABC$: $\angle A = 70°$, $\angle B = 60°$, $\angle C = 50°$. Arrange sides in ascending order.

Smallest $\angle C = 50°$ ⇒ $AB$ (opp. $C$) is shortest.

$\angle B = 60°$ ⇒ $AC$ (opp. $B$) is medium.

Greatest $\angle A = 70°$ ⇒ $BC$ (opp. $A$) is longest.

Ascending order: $AB < AC < BC$.

Worked Example 2 (ML Agarwal Type)

Q. $\triangle ABC$: $BC = 5$, $CA = 6$, $AB = 8$ cm. Find the greatest and least angles without measuring.

Longest side $AB = 8$ cm ⇒ greatest angle = $\angle C$ (opposite $AB$).

Shortest side $BC = 5$ cm ⇒ smallest angle = $\angle A$ (opposite $BC$).

Worked Example 3 — Proving Inequality

Q. Quadrilateral $ABCD$ with diagonal $AC$: $\angle B < \angle A$ and $\angle C < \angle D$. Prove $AD < BC$.

In $\triangle ABC$: $\angle B < \angle A$ ⇒ $AC < BC$ …(i)

In $\triangle ACD$: $\angle C < \angle D$ ⇒ $AD < AC$ …(ii)

From (i) and (ii): $AD < AC < BC$, so $AD < BC$. ∴ proved.

Worked Example 4 — Altitude and Inequality

Q. $AD$ is the altitude from $A$ to $BC$. Show $AB > BD$ and $AB > AD$.

In $\triangle ABD$: $\angle ADB = 90°$ (largest angle in the triangle).

The hypotenuse $AB$ (opposite the right angle) is the longest side: $AB > BD$ and $AB > AD$.

✏ Practice Problems — Side-Angle Inequality

$\triangle PQR$: $\angle P = 50°$, $\angle Q = 70°$, $\angle R = 60°$. Arrange $PQ$, $QR$, $PR$ in ascending order.
$\triangle XYZ$: $XY = 4$, $YZ = 7$, $XZ = 5$ cm. Name the greatest and smallest angles.
$\triangle ABC$: $AB = 3$, $BC = 5$, $CA = 4$ cm. Which angle is largest?
$AD \perp BC$ (altitude). Is $AB > BD$? Justify using angle inequality.
$PS$ bisects $\angle P$ in $\triangle PQR$ and $\angle PQR > \angle PRQ$. Prove $PR > PQ$.
$BC$ is produced to $D$ in $\triangle ABC$. Show $\angle ACD > \angle ABC$ and $\angle ACD > \angle BAC$.
$\triangle ABC$ and $\triangle DEF$: $AB = DE$, $BC = EF$, $\angle B > \angle E$. Compare $AC$ and $DF$.
Largest angle in $\triangle = 90°$. Show other two are acute and hypotenuse is longest.
$\angle A > \angle B > \angle C$ in $\triangle ABC$ with opposite sides $a, b, c$. Write $a, b, c$ in descending order.
$O$ inside $\triangle ABC$ with $OB = OC$. Prove $\angle OBC = \angle OCB$. If $\angle OBA < \angle OCA$, what can you say about $O$?

6. Sum of Any Two Sides of a Triangle is Greater than the Third Side

Triangle Inequality Theorem

[Diagram Placeholder]
A clear mathematical illustration. Background color: #fff3e0 (light orange, to match this theorem box). Left side on a #fff3e0 background: triangle ABC with sides a = BC, b = CA, c = AB, and three bold dark-orange inequalities: "a + b > c", "b + c > a", "c + a > b". Right side: geometric demonstration showing a straight dark-blue line B directly to C (direct path = a) alongside red dashed path B-to-P-to-C (indirect route, BP + PC > BC). Label "Direct path (a) is shortest". Background strictly #fff3e0.

Statement: The sum of any two sides of a triangle is strictly greater than the third side.

In $\triangle ABC$: $AB + BC > CA$; $BC + CA > AB$; $CA + AB > BC$.

All three conditions must hold simultaneously.

Proof not required for ICSE examination.

Key Deductions

Validity check: Sum of the two smaller sides must exceed the largest side.
Difference property: $|AB - BC| < CA$ (difference of any two sides < third).
Range of third side: If two sides are $p$ and $q$, third side $r$ satisfies: $|p - q| < r < p + q$.

Worked Example 1 — Validity Check

Q. Which can form a triangle? (a) 3,4,5 (b) 5,8,15 (c) 6,6,11 (d) 1,2,3

(a) $3+4=7>5$ ✓ — Valid.

(b) $5+8=13 \not> 15$ ✗ — Not valid.

(c) $6+6=12>11$ ✓ — Valid.

(d) $1+2=3 \not> 3$ (must be strictly greater) ✗ — Not valid (degenerate).

Worked Example 2 — Finding Range

Q. Two sides of a triangle: 7 cm and 10 cm. Find all possible values for third side $x$.

$x + 7 > 10 \Rightarrow x > 3$.

$7 + 10 > x \Rightarrow x < 17$.

Answer: $3 < x < 17$ cm.

Worked Example 3 — Interior Point

Q. $O$ inside $\triangle ABC$. Prove $OA + OB + OC > \frac{1}{2}(AB + BC + CA)$.

In $\triangle OAB$: $OA + OB > AB$ …(i)

In $\triangle OBC$: $OB + OC > BC$ …(ii)

In $\triangle OCA$: $OC + OA > CA$ …(iii)

Adding: $2(OA+OB+OC) > AB+BC+CA$ ⇒ $OA+OB+OC > \tfrac{1}{2}(AB+BC+CA)$.

Worked Example 4 — Perimeter Property

Q. Prove any side of a triangle is less than half its perimeter.

Sides $a, b, c$; perimeter $s = a+b+c$. By triangle inequality: $b+c > a$.

Adding $a$: $s > 2a \Rightarrow a < \tfrac{s}{2}$. Similarly $b < \tfrac{s}{2}$ and $c < \tfrac{s}{2}$.

✏ Practice Problems — Triangle Inequality

State which can be sides of a triangle: (a) 2,3,4 (b) 4,5,10 (c) 1.5,2,3.5 (d) 7,8,9 (e) 0.4,0.3,0.5
Two sides: 5 cm and 12 cm. Find all possible integer values for the third side.
Perimeter = 30 cm, two sides = 10 and 12 cm. Find third side and verify triangle inequality.
Prove any side of a triangle is less than half its perimeter.
$ABCD$ is a quadrilateral. Prove $AB+BC+CD+DA > AC+BD$.
$O$ inside $\triangle PQR$. Prove $PQ+QR+PR > PO+QO+RO$.
Show the difference of any two sides of a triangle is less than the third side.
Two sides are 6 cm and 4 cm. What are all possible values of the third side $x$?
$P$ is any point on side $BC$ of $\triangle ABC$. Prove $AB + AC > 2AP$.
Three sides are consecutive integers and perimeter is between 24 and 30. Find all possible triangles.

7. The Perpendicular is the Shortest Distance from an External Point to a Line

Theorem 3 — Shortest Distance Theorem

[Diagram Placeholder]
A clean mathematical diagram. Background color: #fff3e0 (light orange, to match this theorem box). Horizontal black line l at the bottom on a #fff3e0 background. Point P above the line. From P draw: (1) a vertical dashed blue line perpendicular to l at M with a right-angle square at M, labelled "PM = perpendicular (shortest)"; (2) three oblique red dashed lines to points N1, N2, N3 on the line labelled "PN1, PN2, PN3 (all longer)". Annotation: "PM less than PN for all N ≠ M". Background strictly #fff3e0.

Statement: Of all straight lines drawn from a given external point to a given line, the perpendicular is the shortest.

$P$ is not on line $l$. $PM \perp l$ (foot at $M$). For any $N \neq M$ on $l$:

$PM < PN$

Proof not required for ICSE examination.

Intuitive Explanation

In $\triangle PMN$ where $PM \perp l$: $\angle PMN = 90°$ is the largest angle. By the angle-side inequality, the side opposite the largest angle (hypotenuse $PN$) is the longest. So $PN > PM$. This holds for every $N \neq M$ on the line.

Worked Example 1

Q. $PM \perp AB$, $N$ any other point on $AB$. Prove $PM < PN$.

In $\triangle PMN$: $\angle PMN = 90°$ (largest angle).

Hypotenuse $PN$ (opposite 90°) is the longest side: $PN > PM$ and $PN > MN$.

Therefore $PM < PN$.

Worked Example 2 — Altitude Application

Q. Right $\triangle ABC$, $\angle C = 90°$, $CD \perp AB$ at $D$. Prove $CD < CA$ and $CD < CB$.

In $\triangle ACD$: $\angle CDA = 90°$ ⇒ hypotenuse $AC > CD$, so $CD < CA$.

In $\triangle BCD$: $\angle CDB = 90°$ ⇒ hypotenuse $BC > CD$, so $CD < CB$.

Worked Example 3

Q. A student stands 5 m from a wall (perpendicular). A friend walks diagonally to the wall (7 m path). Which is shorter and by how much?

Perpendicular path = 5 m; oblique path = 7 m.

Perpendicular is shorter by $7 - 5 = 2$ m.

✏ Practice Problems — Shortest Distance and Perpendicular

$PM \perp l$, $N$ and $K$ are other points on $l$. State and justify whether $PM < PN$ and $PM < PK$.
Prove the altitude from any vertex of a triangle is shorter than either of the two sides containing that vertex.
$AD \perp BC$ in $\triangle ABC$. Prove $AB > BD$ and $AB > AD$.
A person at $P$ wants to reach a straight road. Which direction minimizes distance? Explain using the theorem.
$O$ is centre of circle, $OM \perp$ chord $AB$ at $M$. Prove $OM < OA$. What does this say about the distance from centre to chord?
Right $\triangle ABC$, $\angle C = 90°$, $CD \perp AB$. Prove $CD < CA$ and $CD < CB$. Which among $CA$, $CB$, $CD$ is shortest?
$H$ is the orthocentre of $\triangle ABC$. Altitude from $A$ is $AD$. Prove $AD < AB$ and $AD < AC$.
$O$ is the centre of a circle, $AB$ is a chord, $OM \perp AB$ at $M$ (midpoint of $AB$). Prove $OM < OA$ and hence $OM$ is less than the radius.

8. Mixed Applications and Higher Order Problems

Mixed Example 1

Q. $\triangle ABC$ with $AB = AC$. $D$ on $BC$. Prove $AB > AD$.

$D$ on $BC$ ⇒ $\angle ADB$ is exterior angle of $\triangle ADC$: $\angle ADB > \angle ACD = \angle ABC$ (since $AB = AC$).

In $\triangle ABD$: $\angle ADB > \angle ABD$ ⇒ $AB > AD$ (side opp. larger angle is larger).

Mixed Example 2 — Median Inequality

Q. $D$ is midpoint of $BC$ in $\triangle ABC$. Prove $AB + AC > 2AD$.

Produce $AD$ to $E$ with $DE = AD$. Join $EC$.

In $\triangle ABD$ and $\triangle ECD$: $AD = ED$, $BD = CD$ (midpoint), $\angle ADB = \angle EDC$ (V.O.A.).
By SAS: $\triangle ABD \cong \triangle ECD$. By CPCT: $AB = EC$.

In $\triangle AEC$: $AE < AC + CE$ ⇒ $2AD < AC + AB$ ⇒ $AB + AC > 2AD$.

Mixed Example 3 — Parallelogram

Q. $ABCD$ is a parallelogram. Prove $\triangle ABC \cong \triangle CDA$.

In $\triangle ABC$ and $\triangle CDA$: $AB = CD$ (opp. sides), $BC = DA$ (opp. sides), $AC = CA$ (common).

By SSS: $\triangle ABC \cong \triangle CDA$. By CPCT: $\angle ABC = \angle CDA$ (opposite angles equal).

✏ Grand Mixed Practice — Examination Level (ML Agarwal Type)

$T$ on $QR$ of $\triangle PQR$, $S$ is a point with $RT = ST$. Prove $PQ + PR > PS$.
Altitude $AD$ from $A$ in $\triangle ABC$. Prove: (a) $AB > BD$ and (b) $AC > DC$.
Isosceles $\triangle ABC$ ($AB = AC$). Bisectors of $\angle B$ and $\angle C$ meet at $I$. Prove $\triangle BIC$ is isosceles. If $\angle A = 50°$, find $\angle BIC$.
$ABCD$ is a rectangle. Prove both diagonals are equal using congruence.
$\triangle PQR$: $PQ > QR$. $S$ is midpoint of $PR$. Prove $\angle PQS < \angle RQS$.
In $\triangle ABC$ and $\triangle PQR$: $AB = PQ$, $BC = QR$, median $BM$ to $AC$ = median $QN$ to $PR$. Prove $\triangle ABC \cong \triangle PQR$.
$\angle B = \angle E$, $AB = DE$, $F$ midpoint of $BE$. Prove $\triangle ABF \cong \triangle DEF$ and $AF = DF$.
Prove: if in two right triangles, the hypotenuse and one acute angle of one equal those of the other, the triangles are congruent. (Use AAS.)
$O$ is inside regular hexagon $ABCDEF$. Prove $OA+OB+OC+OD+OE+OF >$ half the perimeter.
Isosceles $\triangle ABC$: $AB = AC = 5$ cm, $BC = 6$ cm. Altitude from $A$ meets $BC$ at $D$. (i) Find $BD$ and $AD$. (ii) Verify $AD < AB$. (iii) Verify $AB + AC > 2AD$.
$\triangle ABC$: $AB = AC$. Point $D$ inside with $\angle DBC = \angle DCB$. Prove $AD$ bisects $\angle BAC$.
Right $\triangle PQR$, right angle at $Q$, $QS \perp PR$ at $S$. Compare $QS$, $PQ$, $QR$, and $PR$. State which is shortest and why.

9. Chapter Summary — Quick Revision Card

Key Results at a Glance

#	Result	Statement (Short Form)
1	SSS	3 sides equal ⇒ triangles congruent
2	SAS	2 sides + included angle equal ⇒ congruent
3	AAS	2 angles + any 1 side equal ⇒ congruent
4	RHS	Right angle + hypotenuse + 1 side (right triangles only)
5	Isosceles Theorem	$AB = AC \Rightarrow \angle B = \angle C$ and converse
6	Side-Angle Inequality	Greater side ⇔ greater opposite angle (and converse)
7	Triangle Inequality	Sum of any 2 sides > third; difference of any 2 sides < third
8	Shortest Distance	Perpendicular from external point to line is shortest
9	CPCT	After congruence proved, all corresponding parts are equal
10	AAA / SSA	Do NOT prove congruence; AAA proves similarity only

Common Mistakes to Avoid

Order in congruence statement: $\triangle ABC \cong \triangle DEF$ means $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$ strictly.
SSA is invalid (except RHS): non-included angle gives ambiguous case.
Check ALL three inequalities for triangle validity; all must hold simultaneously.
Hypotenuse in RHS = side opposite the right angle (always the longest side).
Do not use AAA for congruence: it gives similarity only.
CPCT requires congruence to be established first; cannot use it as a reason before proving congruence.

Mnemonic: "Some Strong Students Are Really Hard-working" = SSS, SAS, AAS, RHS. AAA and SSA do NOT prove congruence!

Step-by-Step Approach for Any Congruence Problem

Read the given information and mark it on the figure.
Identify the two triangles to be proved congruent.
Determine the vertex correspondence between the two triangles.
List three matching elements (sides/angles) with reasons (Given / CPCT / V.O.A. / common side etc.).
State the congruence rule (SSS / SAS / AAS / RHS).
Write the congruence statement with correct vertex order.
Use CPCT to state further conclusions.

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