Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle. (Figure 5.24)

Question

Philoid · Accepted Answer

Given ABCD is a parallelogram

AR bisects ∠BAD, DP bisects ∠ADC , CP bisects ∠BCD and BR bisects ∠CBA

∠BAD + ∠ABC = 180° (adjacent ∠s of parallelogram are supplementary)

But 1/2 ∠BAD = ∠BAR

1/2 ∠ABC = ∠RBA

∠BAR + ∠RBA = 1/2 × 180° = 90°

⇒ Δ ARB is right angled at ∠ R since its acute interior angles are complementary.

Similarly Δ DPC is right angled at ∠ P and

Also in Δ COB , ∠BOC = 90° ⇒ ∠POR = 90° (vertically opposite angles)

Similarly in ΔADS , ∠ASD = 90° = ∠PSR (vertically opposite angles)

Since vertically opposite angles are equal and measures 90° the quadrilateral is a rectangle.