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 = ${\displaystyle \frac{1}{2}}$ area of || gm ABCD

(b) Area of APQD = Area of BPQC

Answer 1

(a) A diagonal divides a parallelogram into two triangles of equal areas.

⇒ A(ΔADB) = ${\displaystyle \frac{1}{2}}$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 = ${\displaystyle \frac{1}{2}}$A(|| gm ABCD)  [From (i)] ....(ii)

(b) A(ΔABC) = ${\displaystyle \frac{1}{2}}$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 = ${\displaystyle \frac{1}{2}}$A(|| gm ABCD)   ....(iii)

∴ Area of APQD = Area of BPQC.  ...[From (ii) and (iii)]