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

Mid-Point Theorem

Syllabus at a Glance

ICSE Class 9 – Chapter 10: Mid-Point Theorem

Mid-Point Theorem: Proof and simple applications.
Converse of Mid-Point Theorem: Proof and simple applications.
Equal Intercept Theorem: Proof and simple applications.

1. The Mid-Point Theorem

Theorem 1 — Mid-Point Theorem

[Diagram Placeholder]
A precise mathematical diagram. Background color: #fff3e0 (light orange). A triangle ABC drawn on a #fff3e0 background. Points D and E are the mid-points of sides AB and AC respectively. AD and DB have single red ticks. AE and EC have double blue ticks. The line segment DE is drawn and marked parallel to BC with a green arrow on DE and a green arrow on BC. Bold label: "DE || BC and DE = 1/2 BC". Background strictly #fff3e0.

Statement: The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

If $D$ and $E$ are the mid-points of $AB$ and $AC$ respectively in $\triangle ABC$, then:

$$DE \parallel BC \quad \text{and} \quad DE = \frac{1}{2}BC$$

Proof of Mid-Point Theorem (ICSE Required)

[Diagram Placeholder]
A precise mathematical proof diagram. Background color: #ede7f6 (light purple). A triangle ABC on a #ede7f6 background. D and E are midpoints of AB and AC. Line DE is extended to a point F such that DE = EF. A line is drawn from C to F, which is parallel to AB. The resulting quadrilateral DBCF is formed. Angle A and angle ECF are marked as equal alternate interior angles. Triangles ADE and CFE are lightly shaded to show they are congruent. Background strictly #ede7f6.

Given: $\triangle ABC$ in which $D$ is the mid-point of $AB$ ($AD = DB$) and $E$ is the mid-point of $AC$ ($AE = EC$).

To Prove: $DE \parallel BC$ and $DE = \frac{1}{2}BC$.

Construction: Extend line segment $DE$ to a point $F$ such that $DE = EF$. Join $C$ to $F$.

Proof:

Step 1: In $\triangle ADE$ and $\triangle CFE$:
$AE = CE$ (Given, $E$ is the mid-point of $AC$) …(i)
$\angle AED = \angle CEF$ (Vertically opposite angles) …(ii)
$DE = FE$ (By construction) …(iii)
Therefore, by SAS congruence rule, $\triangle ADE \cong \triangle CFE$.

Step 2: By CPCT (Corresponding Parts of Congruent Triangles):
$AD = CF$ …(iv)
$\angle DAE = \angle FCE$ …(v)

Step 3: But $\angle DAE$ and $\angle FCE$ are alternate interior angles made by transversal $AC$ intersecting lines $AB$ and $CF$.
Since alternate interior angles are equal, $AB \parallel CF \Rightarrow DB \parallel CF$.

Step 4: Also, $AD = DB$ (Given). From (iv), $AD = CF$. Therefore, $DB = CF$.

Step 5: In quadrilateral $DBCF$, one pair of opposite sides ($DB$ and $CF$) are both equal and parallel.
Therefore, $DBCF$ is a parallelogram.

Step 6: Since $DBCF$ is a parallelogram, its opposite sides are parallel and equal.
Thus, $DF \parallel BC \Rightarrow \mathbf{DE \parallel BC}$.

Step 7: Also, $DF = BC$. But $DF = DE + EF = 2DE$ (since $DE = EF$).
$2DE = BC \Rightarrow \mathbf{DE = \frac{1}{2}BC}$. Hence Proved.

Worked Example 1 — Direct Application

Q. In $\triangle PQR$, $M$ and $N$ are mid-points of sides $PQ$ and $PR$ respectively. If $MN = 6$ cm, find the length of $QR$. What can you say about $MN$ and $QR$?

Since $M$ and $N$ are mid-points of $PQ$ and $PR$, by the Mid-Point Theorem, $MN \parallel QR$ and $MN = \frac{1}{2}QR$.

$QR = 2 \times MN = 2 \times 6 = 12$ cm.

Answer: $QR = 12$ cm, and $MN$ is parallel to $QR$.

Worked Example 2 — Perimeter of Inner Triangle

Q. $D, E, F$ are mid-points of sides $BC, CA, AB$ respectively of $\triangle ABC$. If $AB = 6$ cm, $BC = 8$ cm, $CA = 10$ cm, find the perimeter of $\triangle DEF$.

By Mid-Point Theorem:

$DE$ joins mid-points of $AC$ and $BC$, so $DE = \frac{1}{2}AB = \frac{1}{2}(6) = 3$ cm.

$EF$ joins mid-points of $AB$ and $AC$, so $EF = \frac{1}{2}BC = \frac{1}{2}(8) = 4$ cm.

$DF$ joins mid-points of $AB$ and $BC$, so $DF = \frac{1}{2}AC = \frac{1}{2}(10) = 5$ cm.

Perimeter of $\triangle DEF = DE + EF + DF = 3 + 4 + 5 = 12$ cm.

The perimeter of the triangle formed by joining the mid-points is exactly half the perimeter of the original triangle.

Worked Example 3 — Quadrilateral Application

Q. Show that the quadrilateral formed by joining the mid-points of the consecutive sides of any quadrilateral is a parallelogram.

Let $ABCD$ be a quadrilateral and $P, Q, R, S$ be mid-points of $AB, BC, CD, DA$. Join diagonal $AC$.

In $\triangle ABC$: $P, Q$ are mid-points of $AB, BC$. By MPT: $PQ \parallel AC$ and $PQ = \frac{1}{2}AC$. …(i)

In $\triangle ADC$: $S, R$ are mid-points of $AD, CD$. By MPT: $SR \parallel AC$ and $SR = \frac{1}{2}AC$. …(ii)

From (i) and (ii): $PQ \parallel SR$ and $PQ = SR$.

In quadrilateral $PQRS$, one pair of opposite sides is equal and parallel. Therefore, $PQRS$ is a parallelogram. Proved.

2. Converse of Mid-Point Theorem

Theorem 2 — Converse of Mid-Point Theorem

[Diagram Placeholder]
A precise mathematical diagram. Background color: #fff3e0 (light orange). A triangle ABC on a #fff3e0 background. D is the mid-point of AB (AD = DB marked with red ticks). A line DE is drawn parallel to BC (indicated by arrows on DE and BC). The point E lies on AC. Bold label: "Line through mid-point D, parallel to BC, bisects AC at E (AE = EC)". Background strictly #fff3e0.

Statement: The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

In $\triangle ABC$: If $D$ is the mid-point of $AB$ AND $DE \parallel BC$, then $E$ is the mid-point of $AC$.

Proof of Converse of Mid-Point Theorem

[Diagram Placeholder]
A precise mathematical proof diagram. Background color: #ede7f6 (light purple). A triangle ABC on a #ede7f6 background. D is the mid-point of AB. Line DE is drawn parallel to BC, intersecting AC at E. A line is drawn from C parallel to AB, meeting the extension of DE at point F. Quadrilateral DBCF is formed. Angles are marked to show congruent triangles ADE and CFE. Background strictly #ede7f6.

Given: In $\triangle ABC$, $D$ is the mid-point of $AB$ ($AD = DB$). A line through $D$ is parallel to $BC$ ($DE \parallel BC$) and intersects $AC$ at $E$.

To Prove: $AE = EC$ (i.e., $E$ is the mid-point of $AC$).

Construction: Draw a line through $C$ parallel to $AB$, meeting $DE$ produced at $F$.

Proof:

Step 1: Since $DF \parallel BC$ (Given) and $DB \parallel CF$ (By construction).
Quadrilateral $DBCF$ is a parallelogram.

Step 2: Opposite sides of a parallelogram are equal: $DB = CF$.
But $AD = DB$ (Given). Therefore, $AD = CF$.

Step 3: In $\triangle ADE$ and $\triangle CFE$:
$\angle DAE = \angle FCE$ (Alternate interior angles, since $AB \parallel CF$) …(i)
$\angle ADE = \angle CFE$ (Alternate interior angles, since $AB \parallel CF$) …(ii)
$AD = CF$ (Proved in Step 2) …(iii)

By ASA congruence rule, $\triangle ADE \cong \triangle CFE$.

Step 4: By CPCT, $AE = CE$.
Therefore, $E$ is the mid-point of $AC$. Hence Proved.

Worked Example 4 — Application of Converse

Q. In $\triangle ABC$, $D$ is the mid-point of $AB$. A line is drawn through $D$ parallel to $BC$ meets $AC$ at $E$. If $AC = 14$ cm, find $AE$.

By the Converse of Mid-Point Theorem, since $D$ is the mid-point of $AB$ and $DE \parallel BC$, $E$ must be the mid-point of $AC$.

Therefore, $AE = \frac{1}{2}AC = \frac{1}{2} \times 14 = 7$ cm.

Worked Example 5 — Medians and Centroid

Q. Prove that the line segment joining the mid-points of two sides of a triangle bisects the median drawn to the third side.

Let $ABC$ be a triangle. $D, E$ are midpoints of $AB, AC$. $AM$ is the median to $BC$ (so $M$ is midpoint of $BC$). Let $AM$ intersect $DE$ at $O$.

In $\triangle ABM$, $D$ is the mid-point of $AB$. By Mid-Point Theorem on $\triangle ABC$, $DE \parallel BC$, which means $DO \parallel BM$.

In $\triangle ABM$, $D$ is mid-point of $AB$ and $DO \parallel BM$. By the Converse of Mid-Point Theorem, $O$ is the mid-point of $AM$.

Therefore, $DE$ bisects $AM$. Proved.

✏ Practice Problems — Mid-Point Theorem & Converse

In $\triangle ABC$, $P$ and $Q$ are mid-points of $AB$ and $AC$. If $BC = 7.2$ cm, what is the length of $PQ$?
In $\triangle PQR$, $X, Y, Z$ are mid-points of $PQ, QR, RP$. If $PQ=8, QR=9, RP=11$, find the perimeter of quadrilateral $PXYZ$.
$ABCD$ is a rhombus. Prove that the quadrilateral formed by joining the mid-points of its sides is a rectangle.
In $\triangle ABC$, $AD$ is the median. A line through $D$ parallel to $AB$ meets $AC$ at $E$. Prove that $E$ is the mid-point of $AC$.
$ABCD$ is a trapezium with $AB \parallel DC$. $E$ is the mid-point of $AD$. A line through $E$ parallel to $AB$ meets $BC$ at $F$. Prove that $F$ is the mid-point of $BC$.

3. Equal Intercept Theorem

What is an Intercept?

When a line (called a transversal) intersects two or more other lines, the line segment cut off by these lines on the transversal is called an intercept.

If transversal $p$ cuts lines $l$ and $m$ at points $A$ and $B$, then $AB$ is the intercept made by $l$ and $m$ on $p$.

Theorem 3 — Equal Intercept Theorem

[Diagram Placeholder]
A precise mathematical diagram. Background color: #fff3e0 (light orange). Three parallel lines l, m, n on a #fff3e0 background. A transversal p intersects them at points A, B, C respectively. A second transversal q intersects them at points D, E, F respectively. The intercept AB is marked equal to intercept BC (AB = BC). Bold label indicates: "If AB = BC, then DE = EF". Background strictly #fff3e0.

Statement: If a transversal makes equal intercepts on three or more parallel lines, then any other transversal intersecting them will also make equal intercepts.

If lines $l \parallel m \parallel n$ and transversal $p$ cuts them at $A, B, C$ such that $AB = BC$, then for any other transversal $q$ cutting them at $D, E, F$, we will have $DE = EF$.

Proof of Equal Intercept Theorem

[Diagram Placeholder]
A precise mathematical proof diagram. Background color: #ede7f6 (light purple). Three parallel lines l, m, n on a #ede7f6 background. Transversals p (cuts at A,B,C) and q (cuts at D,E,F). Given AB = BC. A line segment is drawn from A parallel to q, cutting m at G and n at H. Triangles ABG and BCH are formed, or a line through E parallel to p cutting l at G and n at H. Show the construction lines used to prove DE = EF using congruent triangles. Background strictly #ede7f6.

Given: Three parallel lines $l, m, n$. Transversal $p$ intersects them at $A, B, C$ respectively such that $AB = BC$. Another transversal $q$ intersects them at $D, E, F$.

To Prove: $DE = EF$.

Construction: Through point $E$, draw a line parallel to transversal $p$, intersecting line $l$ at $G$ and line $n$ at $H$.

Proof:

Step 1: In quadrilateral $ABEG$:
$AB \parallel GE$ (By construction)
$AG \parallel BE$ (Since $l \parallel m$)
Therefore, $ABEG$ is a parallelogram. $\Rightarrow \mathbf{AB = GE}$.

Step 2: In quadrilateral $BCHE$:
$BC \parallel EH$ (By construction)
$BE \parallel CH$ (Since $m \parallel n$)
Therefore, $BCHE$ is a parallelogram. $\Rightarrow \mathbf{BC = EH}$.

Step 3: We are given $AB = BC$.
From Step 1 and 2, it follows that $\mathbf{GE = EH}$.

Step 4: In $\triangle GDE$ and $\triangle HFE$:
$\angle GED = \angle HEF$ (Vertically opposite angles)
$\angle EGD = \angle EHF$ (Alternate interior angles, since $l \parallel n$ and $GH$ is transversal)
$GE = EH$ (Proved in Step 3)

By ASA congruence rule, $\triangle GDE \cong \triangle HFE$.

Step 5: By CPCT, $DE = FE$, which means $DE = EF$. Hence Proved.

Worked Example 6 — Division of Line Segment

Q. Line segment $AB$ of length 12 cm is to be divided into 3 equal parts. Explain how Equal Intercept Theorem justifies the geometric construction method.

To divide $AB$ into 3 equal parts, we draw a ray $AX$ at an acute angle to $AB$.

We mark 3 equal segments $AA_1 = A_1A_2 = A_2A_3$ on $AX$. Join $A_3$ to $B$.

Draw lines through $A_1$ and $A_2$ parallel to $A_3B$, meeting $AB$ at $P$ and $Q$.

Here, the parallel lines $A_1P \parallel A_2Q \parallel A_3B$ are cut by transversal $AX$ making equal intercepts ($AA_1 = A_1A_2 = A_2A_3$).

By the Equal Intercept Theorem, they must make equal intercepts on the other transversal $AB$. Thus, $AP = PQ = QB$.

Since $AB = 12$ cm, each part is $12 \div 3 = 4$ cm.

✏ Practice Problems — Equal Intercept Theorem

Lines $l \parallel m \parallel n$. A transversal cuts them such that the intercepts are 4 cm and 4 cm. Another transversal cuts them, and the first intercept is 5.5 cm. Find the length of the second intercept.
In $\triangle ABC$, $M, N, P$ are points on $AB$ such that $AM = MN = NB$. Lines through $M$ and $N$ parallel to $BC$ meet $AC$ at $X$ and $Y$. Prove $AX = XY = YC$.
$ABCD$ is a parallelogram. A line through $A$ cuts $DC$ at $P$ and $BC$ produced at $Q$. Prove that if $P$ is the mid-point of $DC$, then $C$ is the mid-point of $BQ$.

4. Chapter Summary — Quick Revision Card

Key Results at a Glance

#	Theorem / Property	Core Concept
1	Mid-Point Theorem	Line joining mid-points of 2 sides is $\parallel$ to 3rd side and = $\frac{1}{2}$ of it.
2	Converse of M.P.T.	Line through mid-point $\parallel$ to one side bisects the 3rd side.
3	Mid-Point Triangle	Perimeter of triangle formed by 3 mid-points = $\frac{1}{2}$ original perimeter.
4	Quadrilateral Mid-Points	Joining mid-points of ANY quadrilateral forms a Parallelogram.
5	Rectangle Mid-Points	Joining mid-points of a rectangle forms a Rhombus.
6	Rhombus Mid-Points	Joining mid-points of a rhombus forms a Rectangle.
7	Square Mid-Points	Joining mid-points of a square forms a Square.
8	Equal Intercept Thm.	Equal intercepts on one transversal $\Rightarrow$ equal intercepts on ALL transversals across the same parallel lines.

Common Pitfalls to Avoid

Do not apply the Mid-Point Theorem if only ONE mid-point is given (unless you are using the Converse and have parallel lines).
In proofs, always state the theorem explicitly: "By Mid-Point Theorem" or "By Converse of Mid-Point Theorem".
Remember that the line segment is not just parallel, it is exactly half the length of the third side.
For the Equal Intercept Theorem, you MUST have at least three parallel lines. It does not apply to just two lines.

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