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Question
The diagonals of a parallelogram ABCD intersect at O. A line through O meets AB in P and CD in Q. Show that
(a) Area of APQD = `(1)/(2)` area of || gm ABCD
(b) Area of APQD = Area of BPQC
Solution
(a) A diagonal divides a parallelogram into two triangles of equal areas.
⇒ A(ΔADB) = `(1)/(2)`A(|| gm ABCD) ....(i)
In ΔDOQ and ΔBOP,
∠DOQ = ∠BOP ...(Vertically opposite angles)
DO = BO ...(Diagonals of a || gm bisect each other)
∠ODQ = ∠OBP ...(Alternate angles)
∴ ΔDOQ ≅ ΔBOP ...(ASA test of congruency)
⇒ A(ΔDOQ) = A(ΔBOP)
Adding A(DOQ) on both sides, we get
A(ΔDOQ) + A(DOPA) = A(ΔBOP) + A(DOPA)
⇒ Area of APQD = A(ADB)
⇒ Area of APQD = `(1)/(2)`A(|| gm ABCD) [From (i)] ....(ii)
(b) A(ΔABC) = `(1)/(2)`A(|| gm ABCD)
In ΔCOQ and ΔAOP,
∠COQ = ∠AOP ...(Vertically opposite angles)
CCO = AO ...(Diagonals of a || gm bisect each other)
∠OCQ = ∠OAP ...(Alternate angles)
∴ ΔCOQ ≅ ΔAOP ...(ASA test of congruency)
⇒ A(ΔCOQ) = A(ΔAOP)
Adding A(COPB) on both sides, we get
A(ΔCOQ) + A(COPB) = A(ΔAOP) + A(COPB)
∴ Area of BPQC = A(ABC)
⇒ Area of BPQC = `(1)/(2)`A(|| gm ABCD) ....(iii)
∴ Area of APQD = Area of BPQC. ...[From (ii) and (iii)]
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