Triangles

(a) \(\Delta ABC \sim \Delta PQR \Rightarrow \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}\).
\(\frac{6}{PQ} = \frac{4}{QR} = \frac{8}{6} = \frac{4}{3}\).
\(PQ = 6 \times \frac{3}{4} = 4.5\).
\(QR = 4 \times \frac{3}{4} = 3\).
\(PQ + QR = 4.5 + 3 = 7.5\).

(b) Given \(\angle 1 = \angle 2 \Rightarrow PQ = PR\) (Isosceles \(\Delta PQR\)).
Substitute \(PR = PQ\) in given ratio: \(\frac{QR}{QS} = \frac{QT}{PQ}\).
In \(\Delta PQS\) and \(\Delta TQR\):
\(\frac{QS}{QR} = \frac{PQ}{QT}\) (Reciprocal of above).
\(\angle Q\) is common.
By SAS similarity, \(\Delta PQS \sim \Delta TQR\).

(a) In parallelogram ABCD, \(AD || BC\).
\(\Delta AFD \sim \Delta BFE\) (AA Similarity: Alt Int Angles).
\(\frac{DF}{FB} = \frac{FA}{EF} \Rightarrow DF \times EF = FB \times FA\).

(b) Given \(AD^2 = BD \times DC \Rightarrow \frac{AD}{BD} = \frac{DC}{AD}\).
In \(\Delta ADB\) and \(\Delta CDA\):
\(\angle ADB = \angle CDA = 90^\circ\).
Sides including these angles are proportional.
So \(\Delta ADB \sim \Delta CDA\) (SAS).
Let \(\angle ABD = x \Rightarrow \angle BAD = 90 - x\).
From similarity, \(\angle CAD = \angle ABD = x\).
\(\angle BAC = \angle BAD + \angle CAD = (90 - x) + x = 90^\circ\).

By Basic Proportionality Theorem (Thales Theorem), since \(PQ || BC\):
\(\frac{AP}{PB} = \frac{AQ}{QC}\).
\(\frac{AQ}{QC} = \frac{4}{13} \Rightarrow QC = \frac{13}{4}AQ\).
\(AC = AQ + QC = AQ + \frac{13}{4}AQ = \frac{17}{4}AQ\).
Given \(AC = 20.4\).
\(20.4 = \frac{17}{4}AQ \Rightarrow AQ = \frac{20.4 \times 4}{17} = 1.2 \times 4 = 4.8\) cm.
Answer: (D)

(A) True. Congruent figures have same shape and size, so they are similar.
(B) False. A square has all angles 90°. A rhombus may have any angles. Even with same area, their angles (shape) differ. So not similar.
(C) True. Equilateral triangles always have 60° angles.
(D) True. Similar figures can have different sizes.
Answer: (B)

(a)
Given \(\frac{PS}{SQ} = \frac{PT}{TR}\). By Converse of BPT, \(ST || QR\).
Therefore, corresponding angles area equal: \(\angle PST = \angle PQR\).
Given \(\angle PST = \angle PRQ\).
So, \(\angle PQR = \angle PRQ\).
Sides opposite to equal angles are equal: \(PQ = PR\).
Hence, \(\Delta PQR\) is an isosceles triangle.

(b)
Given \(\Delta ABE \cong \Delta ACD\).
By CPCT, \(AB = AC\) and \(AE = AD\).
\(\frac{AB}{AC} = 1\) and \(\frac{AD}{AE} = 1\).
Therefore, \(\frac{AD}{AB} = \frac{AE}{AC}\).
In \(\Delta ADE\) and \(\Delta ABC\):
1. \(\frac{AD}{AB} = \frac{AE}{AC}\) (Proved above)
2. \(\angle A = \angle A\) (Common)
By SAS Similarity Criterion, \(\Delta ADE \sim \Delta ABC\).

Part 1: Basic Proportionality Theorem (Thales Theorem)
Statement: If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
Proof: Given \(\Delta ABC\) and line \(DE \parallel BC\) intersecting \(AB\) at \(D\) and \(AC\) at \(E\).
Draw \(EM \perp AB\) and \(DN \perp AC\). Join \(BE\) and \(CD\).
Area(\(\Delta ADE\)) = \(\frac{1}{2} \times AD \times EM\).
Area(\(\Delta BDE\)) = \(\frac{1}{2} \times DB \times EM\).
Ratio \(\frac{Area(\Delta ADE)}{Area(\Delta BDE)} = \frac{AD}{DB}\). ... (i)
Similarly, \(\frac{Area(\Delta ADE)}{Area(\Delta DEC)} = \frac{AE}{EC}\). ... (ii)
Since \(\Delta BDE\) and \(\Delta DEC\) are on the same base \(DE\) and between same parallel lines \(DE\) and \(BC\), Area(\(\Delta BDE\)) = Area(\(\Delta DEC\)).
Therefore, from (i) and (ii), \(\frac{AD}{DB} = \frac{AE}{EC}\). Proved.

Part 2:
In \(\Delta ABC\), given \(LM || CB\).
By BPT, \(\frac{AM}{MB} = \frac{AL}{LC}\) --- (1)
In \(\Delta ADC\), given \(LN || CD\).
By BPT, \(\frac{AN}{ND} = \frac{AL}{LC}\) --- (2)
From (1) and (2):
\(\frac{AM}{MB} = \frac{AN}{ND}\). Proved.

Correspondence: \(D \leftrightarrow Q, E \leftrightarrow R, F \leftrightarrow P\).
\(\Delta DEF \sim \Delta QRP\).
Ratios: \(\frac{DE}{QR} = \frac{EF}{RP} = \frac{DF}{QP}\).
(A) Correct. (B) Correct. (C) Correct.
(D) \(\frac{DE}{PQ}\) compares side DE with PQ (which is QP's partner is DF). Incorrect.
Answer: (D)

By similarity or Geometric Mean theorem in right triangle: \(AC^2 = CD \times CB\).
\(AC^2 = 4 \times 16 = 64\).
\(AC = 8\) cm.
Answer: (C)

Observations from Figure:
In \(\Delta LMN\), side lengths are \(LM=45\), \(MN=63\), \(LN=72\) (from proportionality). \(\angle M = 130^\circ\).
In \(\Delta ABC\), side \(AB=5\), and \(\angle B = 130^\circ\).
Reasoning:
Comparing \(\Delta LMN\) and \(\Delta ABC\):
Ratios of adjacent sides to \(130^\circ\): \(\frac{LM}{AB} = \frac{45}{5} = 9\).
Assume similarity \(\Delta LMN \sim \Delta ABC\) (as angles match and sides are proportional).
The scale factor is \(k = 9\).
The side opposite to \(130^\circ\) in \(\Delta LMN\) is \(LN = 72\).
The side opposite to \(130^\circ\) in \(\Delta ABC\) is \(AC\).
Therefore, \(\frac{LN}{AC} = 9 \Rightarrow \frac{72}{AC} = 9\).
\(AC = \frac{72}{9} = 8\) cm.
Answer: (C)

(a) Let \(AB = a, BC = b\).
In \(\Delta PAC, QB || PA \Rightarrow \frac{y}{x} = \frac{b}{a+b}\).
In \(\Delta ACR, QB || RC \Rightarrow \frac{y}{z} = \frac{a}{a+b}\).
Adding: \(\frac{y}{x} + \frac{y}{z} = \frac{a+b}{a+b} = 1\).
Dividing by \(y\): \(\frac{1}{x} + \frac{1}{z} = \frac{1}{y}\).

(b) Standard theorem proof. Proportionality of sides and median implies similarity (SSS similarity for half triangles by doubling median construction, then SAS for whole).

By BPT, \(EF||BC\). \(\Delta AEF \sim \Delta ABC\).
Ratio \(AE/AB = 1/(1+2) = 1/3\).
So \(EF/BC = 1/3 \Rightarrow BC = 3EF\).
Answer: (d)

\(\angle C = 180 - (80+60) = 40°\).
Answer: (d)

(A):
Given \(\Delta ABC \sim \Delta PQR\) (Implied by ratio of sides).
Ratio of sides = 3 : 5.
(i) Ratio of corresponding altitudes is equal to ratio of sides.
\(\frac{AD}{PS} = \frac{3}{5} \Rightarrow \Delta ADC \sim \Delta PSR\) (Since \(\angle D = \angle S = 90^\circ\) and \(\angle C = \angle R\)).
(ii) \(\frac{AD}{PS} = \frac{3}{5}\). Given \(AD = 4\).
\(\frac{4}{PS} = \frac{3}{5} \Rightarrow PS = \frac{20}{3} \approx 6.67\) cm.
(iii) Ratio of areas = (Ratio of sides)\(^2\) = \((\frac{3}{5})^2 = \frac{9}{25}\).
ar\((\Delta ABC)\) : ar\((\Delta PQR)\) = 9 : 25.

(B): Proof of BPT (Basic Proportionality Theorem):
(See Proof in Q32 30-2 Part 1 above).
Proving Intercept Property:
Join A to F, intersecting line \(m\) at G.
In \(\Delta ACF\), line \(BG \parallel CF\) (given \(m \parallel n\)).
By BPT, \(\frac{AB}{BC} = \frac{AG}{GF}\). ... (i)
In \(\Delta AFD\), line \(GE \parallel AD\) (given \(m \parallel l\)).
By BPT, \(\frac{AG}{GF} = \frac{DE}{EF}\). ... (ii)
From (i) and (ii): \(\frac{AB}{BC} = \frac{DE}{EF}\). Proved.

Right triangles are only similar if corresponding sides are proportional or acute angles are equal. An isosceles right triangle is not similar to a scalene right triangle.
Correct Option: (A)

By the property of parallel lines cutting transversals proportionally: \(\frac{AX}{XD} = \frac{BY}{YC}\).
Given \(AX = \frac{2}{3}AD \Rightarrow AX = 2k, AD = 3k \Rightarrow XD = k\).
Ratio \(\frac{AX}{XD} = \frac{2}{1}\). So \(\frac{BY}{YC} = \frac{2}{1}\).
The question asks for \(CY : YB\), which is the reciprocal \(1 : 2\).
Correct Option: (D) 1:2

From \(\Delta ABD \sim \Delta PQS\): \(\frac{AB}{PQ} = \frac{AD}{PS}\) and \(\angle B = \angle Q\).
From \(\Delta ACD \sim \Delta PRS\): \(\frac{AC}{PR} = \frac{AD}{PS}\) and \(\angle C = \angle R\).
Combining, \(\frac{AB}{PQ} = \frac{AC}{PR}\).
Also, from \(\angle B = \angle Q\) and \(\angle C = \angle R\), \(\angle A = \angle P\) (Angle sum property).
Thus, \(\Delta ABC \sim \Delta PQR\) by AAA (or SSS using altitude ratios).

Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Proof: Since \(AD \parallel CE\) and AC is transversal, \(\angle DAC = \angle ACE\) (Alternate angles).
Since \(AD \parallel CE\) and BE is transversal, \(\angle BAD = \angle AEC\) (Corresponding angles).
Given AD bisects \(\angle A\), so \(\angle BAD = \angle DAC\).
Therefore, \(\angle ACE = \angle AEC\). In \(\Delta ACE\), sides opposite equal angles are equal, so \(AC = AE\).
In \(\Delta BCE\), since \(AD \parallel CE\), by BPT: \(\frac{BD}{DC} = \frac{BA}{AE}\).
Substituting \(AE = AC\), we get \(\frac{BD}{DC} = \frac{BA}{AC}\).

Using BPT (Basic Proportionality Theorem):
In \(\Delta AXY\), \(PQ||XY \Rightarrow \frac{AP}{PX} = \frac{AQ}{QY}\)
\(\frac{2}{1.5} = \frac{AQ}{0.75} \Rightarrow AQ = \frac{2 \times 0.75}{1.5} = 1\) cm.

In \(\Delta ABC\), \(XY||BC \Rightarrow \frac{AX}{XB} = \frac{AY}{YC}\)
\(AX = AP + PX = 2 + 1.5 = 3.5\), \(XB = 4\). \(AY = AQ + QY = 1 + 0.75 = 1.75\).
Using BPT: \(\frac{PX}{XB} = \frac{QY}{YC}\)
\(\frac{1.5}{4} = \frac{0.75}{YC} \Rightarrow YC = \frac{4 \times 0.75}{1.5} = 2\) cm.

Therefore, \(AQ + CY = 1 + 2 = 3\) cm.
Correct Option: (C)

Since \(\Delta ABC \sim \Delta PQR\):
\(\angle A = \angle P = 30°\)
\(\angle B = \angle Q = 90°\)
In \(\Delta PQR\), \(\angle R = 180 - (30 + 90) = 60°\).
Value of \(\angle R + \angle B = 60° + 90° = 150°\).
Correct Option: (C)

\(AB = AP + BP = 1 + 2 = 3\) cm.
Check ratio of sides:
\(\frac{AP}{AB} = \frac{1}{3}\)
\(\frac{AQ}{AC} = \frac{1.5}{4.5} = \frac{1}{3}\)

In \(\Delta APQ\) and \(\Delta ABC\):
1. \(\frac{AP}{AB} = \frac{AQ}{AC}\) (Proved above)
2. \(\angle A = \angle A\) (Common)
By SAS Similarity Criterion, \(\Delta APQ \sim \Delta ABC\).

Finding PQ:
Since triangles are similar, corresponding sides are proportional:
\(\frac{PQ}{BC} = \frac{AP}{AB}\)
\(\frac{PQ}{3.6} = \frac{1}{3}\)
\(PQ = \frac{3.6}{3} = 1.2\) cm.

Solution Q32 (a):
Statement: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Given: A triangle ABC and a line \(l\) intersecting AB at D and AC at E, such that \(\frac{AD}{DB} = \frac{AE}{EC}\).
To Prove: \(DE || BC\).

Proof:
Assume \(DE\) is not parallel to \(BC\).
Then, there must be another line passing through D, say \(DF\), such that \(DF || BC\).
By Basic Proportionality Theorem (Thales Theorem), since \(DF || BC\):
\(\frac{AD}{DB} = \frac{AF}{FC}\) --- (i)
But given \(\frac{AD}{DB} = \frac{AE}{EC}\) --- (ii)
From (i) and (ii), \(\frac{AF}{FC} = \frac{AE}{EC}\).
Adding 1 to both sides: \(\frac{AF}{FC} + 1 = \frac{AE}{EC} + 1\)
\(\frac{AF+FC}{FC} = \frac{AE+EC}{EC}\)
\(\frac{AC}{FC} = \frac{AC}{EC}\)
\(\Rightarrow FC = EC\).
This is only possible if points F and E coincide.
Hence, our assumption was wrong, and \(DE || BC\). Proved.

Solution Q32 (b):
To Prove: \(\Delta ADB \sim \Delta CDA\)
Let \(\angle ABD = x\).
In \(\Delta ADB\) (Right angled at D), \(\angle DAB = 90 - x\).
Since \(\angle A = 90°\), \(\angle CAD = 90 - \angle DAB = 90 - (90 - x) = x\).
In \(\Delta ADB\) and \(\Delta CDA\):
1. \(\angle ADB = \angle CDA = 90°\) (Given)
2. \(\angle ABD = \angle CAD = x\) (Proved)
By AA Similarity, \(\Delta ADB \sim \Delta CDA\).

Find AD:
Since \(\Delta ADB \sim \Delta CDA\):
\(\frac{AD}{CD} = \frac{DB}{DA}\)
\(AD^2 = BD \times CD\)
Given \(BC = 10\) and \(CD = 2\), so \(BD = BC - CD = 10 - 2 = 8\) cm.
\(AD^2 = 8 \times 2 = 16\)
\(AD = 4\) cm.