ICSE Class 9 Maths Chapter 12 Mid Point and Its Converse Including Intercept Theorem

Mid-Point and Its Converse

Including Intercept Theorem

Mid-Point Theorem (Proof and simple applications and its converse)

Theorem 6

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.

Given: D and E are the mid-points of sides AB and AC respectively of triangle ABC.

To Prove: DE is parallel to BC and DE = \(\frac{1}{2}\) BC.

Construction: Draw CF parallel to BA which meets DE produced at F.

Proof

Now, DE = EF (triangle ADE is congruent to triangle CFE)

= \(\frac{1}{2}\) DF

= \(\frac{1}{2}\) BC (In parallelogram BCFD; DF = BC.)

Therefore, DE is parallel to BC and DE = \(\frac{1}{2}\) BC.

Hence Proved

Theorem 7

(Converse of Mid-point Theorem)

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

Given: D is mid-point of side AB of a triangle ABC and DE is drawn parallel to the side BC.

To Prove: DE bisects AC, i.e. AE = EC.

Construction: Draw CF parallel to BA which meets DE produced at F

Proof

Hence Proved

In the figure, given above, D is given to be the mid-point of AB and E is proved to be the mid-point of AC, therefore, DE will be half of the 3rd side i.e., DE = \(\frac{1}{2}\) BC.

Example 1

Prove that the figure obtained by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram.

Solution:

Given: P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively of quadrilateral ABCD.

To Prove: PQRS is a parallelogram.

Construction: Join B and D.

Proof:

1. In triangle ABD:
PS is parallel to BD and PS = \(\frac{1}{2}\) BD (Line joining the mid-points of two sides of a triangle is parallel and half of third side)

2. In triangle BCD:
QR is parallel to BD and QR = \(\frac{1}{2}\) BD (Line joining the mid-points of two sides of a triangle is parallel and half of third side)

Therefore, PS is parallel to QR and PS = QR (From (1) and (2))

Therefore, PQRS is a parallelogram (One pair of opp. sides are equal and parallel)

Hence Proved

Teacher's Note

The mid-point theorem is like finding the center lane of a highway - it divides the road proportionally and creates parallel paths that are half the distance of the original route.

Example 2

In parallelogram PQRS, L is mid-point of side SR and SN is drawn parallel to LQ which meets RQ produced at N and cuts side PQ at M.

Prove that:
(i) SP = \(\frac{1}{2}\) RN
(ii) SN = 2 LQ

Solution:

(i) In triangle SRN:
L is mid-point of SR (Given)
and LQ is parallel to SN (Given)
Therefore, LQ bisects RN (Line through mid-point of one side of a triangle and parallel to another side bisects the third side)

i.e. RQ = QN = \(\frac{1}{2}\) RN
But, SP = RQ (Opposite sides of parallelogram PQRS)
Therefore, SP = \(\frac{1}{2}\) RN

Hence Proved

(ii) In triangle SRN:
L is mid-point of SR (Given)
and Q is mid-point of RN (Proved in part (i))
Therefore, LQ = \(\frac{1}{2}\) SN (Line joining the mid-points of two sides of a triangle is half of the third side)
or, SN = 2 LQ

Hence Proved

Teacher's Note

This example shows how mid-point properties extend to more complex figures - like how GPS navigation divides a complex journey into proportional segments for accurate route planning.

Example 3

The adjoining figure shows a parallelogram ABCD in which P is mid-point of AB and Q is mid-point of CD. Prove that: AE = EF = FC.

Solution:

Since, PB = \(\frac{1}{2}\) AB (Given, P is mid-point of AB)

DQ = \(\frac{1}{2}\) DC (Given, Q is mid-point of DC)

Therefore, PB = DQ (Since, AB = DC; the opp. sides of parallelogram ABCD)

Also, PB is parallel to DQ (As AB is parallel to DC)

Therefore, DPBQ is a parallelogram (Opp. sides PB and DQ are parallel and equal)

Therefore, DP is parallel to QB (Opp. sides of the parallelogram DPBQ)

Now in triangle ABF:
P is mid-point of AB (Given)
PE is parallel to BF (As DP is parallel to QB)
Therefore, PE bisects AF (Line passing through the mid-point of one side and parallel to another bisects the third side)

i.e. AE = EF ........... I

Similarly, in triangle CDE:
QF bisects CE (IQ is mid-point of CD and QF is parallel to DE)

Therefore, EF = FC ........... II

Therefore, AE = EF = FC (From I and II)

Hence Proved

Teacher's Note

This demonstrates how parallelogram properties combined with mid-point theorems create equal divisions - similar to how construction workers divide a wall into equal sections for precise tile placement.

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