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. Prove that ar (ABCD) = ar (APQD).

Explanation

Given In trapezium ABCD, `AB || DC`, DC produced in Q and L is the mid-point of BC.
`therefore" "` BL = CL
To prove `" " ar (ABCD) = ar (APQD)`
Proof Since, DC produced in Q and `AB || DC`.
So, `" " DQ || AB`
In `DeltaCLQ " and " DeltaBLP`,
`CL = BL" "` [since, L is the mid-point of BC]
`angleLCQ = angleLBP" "` [alternate interior angles as BC is a transversal]
`angleCQL = angleLPB" "` [alternate interior angles as PQ is a transversal]
`therefore" "` `DeltaCLQ ~= DeltaBLP" "` [by AAS congruence rule]
Then, `" "ar (DeltaCLQ) = ar (DeltaBLP)" "` ...(i)
[since, congruent triangles have equal area]
Now, `" " ar (ABCD) = ar (APQD) - ar (DeltaCQL) + ar (DeltaBLP)`
`= ar (APQD) - ar (DeltaBLP) + ar (DeltaBLP)" "` [from Eq. (i)]
`rArr" "` `ar (ABCD) = ar (APQD)" "` Hence proved.