Exercise 8.2 Solutions (Class 9)

(i) Show SR || AC and SR = 1/2 AC:
In ΔDAC, S is the mid-point of DA and R is the mid-point of DC.
By the Mid-point Theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it.
Therefore, SR || AC and SR = (1/2)AC. (Proved)

(ii) Show PQ = SR:
In ΔBAC, P is the mid-point of AB and Q is the mid-point of BC.
By the Mid-point Theorem, PQ || AC and PQ = (1/2)AC.
From part (i), we have SR = (1/2)AC. Therefore, PQ = SR. (Proved)

(iii) Show PQRS is a parallelogram:
From the proofs above, we have PQ || AC and SR || AC, which implies PQ || SR.
We also have PQ = SR.
Since one pair of opposite sides (PQ and SR) is both equal and parallel, the quadrilateral PQRS is a parallelogram. (Proved)

Join diagonals AC and BD. From Q1, we know PQRS is a parallelogram.
In ΔADC, SR || AC and SR = 1/2 AC.
In ΔABC, PQ || AC and PQ = 1/2 AC.
In ΔADB, SP || BD and SP = 1/2 BD.
In ΔCDB, RQ || BD and RQ = 1/2 BD.

The diagonals of a rhombus (AC and BD) bisect each other at right angles. So, AC ⊥ BD.

Since SR || AC and RQ || BD, the angle between SR and RQ must be the same as the angle between AC and BD.
Therefore, SR ⊥ RQ, which means ∠SRQ = 90°.

A parallelogram with one right angle is a rectangle. Hence, PQRS is a rectangle. (Proved)

Join diagonals AC and BD. From Q1, we know PQRS is a parallelogram.

In ΔABC, by Mid-point Theorem, PQ = 1/2 AC.
In ΔADC, by Mid-point Theorem, SR = 1/2 AC. So PQ = SR.

In ΔABD, by Mid-point Theorem, SP = 1/2 BD.
In ΔBCD, by Mid-point Theorem, RQ = 1/2 BD. So SP = RQ.

The diagonals of a rectangle are equal, so AC = BD.
Therefore, PQ = SP = RQ = SR (since each is half of the equal diagonals).

A parallelogram with all four sides equal is a rhombus. Hence, PQRS is a rhombus. (Proved)

Given: Trapezium ABCD, AB || DC. E is mid-point of AD. EF || AB.

To Prove: F is the mid-point of BC.

Construction: Let EF intersect diagonal BD at G.

Proof:
In ΔDAB, E is the mid-point of AD and EG || AB (since EF || AB).
By the Converse of the Mid-point Theorem, a line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.
Therefore, G is the mid-point of DB.

Now, in ΔBCD, G is the mid-point of DB (proved above).
Also, since AB || DC and EF || AB, it follows that EF || DC. So, GF || DC.
Again, by the Converse of the Mid-point Theorem, F must be the mid-point of BC.

(Proved)

Since ABCD is a parallelogram, AB || CD and AB = CD.
This means AE || FC and AE = 1/2 AB = 1/2 CD = FC. Since one pair of opposite sides (AE and FC) is equal and parallel, AECF is a parallelogram. Therefore, AF || EC.

In ΔDQC, F is the mid-point of DC and FP || CQ (since AF || EC). By Converse of Mid-point theorem, P is the mid-point of DQ. So, DP = PQ. —(1)

In ΔAPB, E is the mid-point of AB and EQ || AP (since AF || EC). By Converse of Mid-point theorem, Q is the mid-point of PB. So, PQ = QB. —(2)

From (1) and (2), we have DP = PQ = QB. This means the segments AF and EC divide the diagonal BD into three equal parts.

(Proved)

(i) D is the mid-point of AC:
In ΔABC, M is the mid-point of AB and MD || BC. By the Converse of the Mid-point Theorem, D must be the mid-point of the third side AC. (Proved)

(ii) MD ⊥ AC:
Given MD || BC. Since BC ⊥ AC (∠C = 90°), the corresponding angles are equal. Therefore, ∠MDC = ∠BCA = 90°. This means MD ⊥ AC. (Proved)

(iii) CM = MA = 1/2 AB:
Join CM. In ΔADM and ΔCDM:
• AD = CD (D is the mid-point)
• ∠ADM = ∠CDM = 90° (Proved)
• MD = MD (Common)
By SAS congruence, ΔADM ≅ ΔCDM. So, CM = AM (by CPCT).
Since M is the mid-point of AB, AM = 1/2 AB.
Therefore, CM = AM = 1/2 AB. (Proved)