In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (Fig. 9.18). Prove that ar (ABCD) = ar (APQD)

Solution:

Given, ABCD is a trapezium

AB || DC

L is the midpoint of BC

Though L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q.

We have to prove that ar(ABCD) = ar(APQD)

Since L is the midpoint of BC

BL = CL

Since DC is produced in Q and AB || DC

So, DQ || AB

Considering triangles CLQ and BLP,

CL = BL

Since DQ || AB and cut by transversal BC

We know that the alternate interior angles are equal.

∠LCQ = ∠LBP

Since DQ || AB and cut by transversal PQ

∠CQL = ∠LPB

By AAS criterion, the triangles CLQ and BLP are similar.

We know that congruent triangles have equal area

So, ar(CLQ) = ar(BLP) ----------------------- (1)

Now, ar(ABCD) = ar(APQD) - ar(CQL) + ar(BLP)

From (1),

ar(ABCD) = ar(APQD) - ar(BLP) + ar(BLP)

= ar(APQD)

Therefore, ar(ABCD) = ar(APQD)

✦ Try This: In ΔABC, AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are respectively the mid-points of AB and BC, determine the length of DE.

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 9

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NCERT Exemplar Class 9 Maths Exercise 9.3 Problem 8

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (Fig. 9.18). Prove that ar (ABCD) = ar (APQD)

Summary:

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (Fig. 9.18). It is proven that ar (ABCD) = ar (APQD)

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