Question 8
In trapezium ABCD, AB || DC and L is the midpoint of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. prove that ar (ABCD) = ar (APQD).

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Solution

Given,
In trapezium ABCD, AB || DC, DC produced to Q and L is the midpoint of BC.
$∴$ BL = CL
Proof:
DC produced to Q and AB || DC.
So, DQ || AB.
In $Δ C L Q$ and $Δ B L P$ ,
CL = BL [L is the mid point of BC]
$∠ L C Q = ∠ L B P$ [alternate interior angles as BC is a transversal]
$∠ C Q L = ∠ L P B$ [alternate interior angles as PQ is a transversal]
$Δ C L Q ≅ Δ B L P$ [by AAS congruence rule]
Then, $a r (Δ C L Q) = a r (Δ B L P)$ …(i)
Now, $a r (A B C D) = a r (A P Q D) - a r (Δ C Q L) + a r (Δ B L P)$
$= a r (A P Q D) - a r (Δ B L P) + a r (Δ B L P)$ [from Eq. (i)]
$\Rightarrow$ ar (ABCD) = ar (APQD)

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Question 8 In trapezium ABCD, AB || DC and L is the midpoint of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. prove that ar (ABCD) = ar (APQD).

Question 8
In trapezium ABCD, AB || DC and L is the midpoint of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. prove that ar (ABCD) = ar (APQD).