Question

ABCD is a parallelogram. The bisectors of ∠A and ∠B meet |BC and |AD (produced if necessary) at X and Y respectively.
Prove XY = CD

Answer 1

ABCD is a parallelogram. The bisectors of $\angle A$ and $\angle B$ meet |BC and |AD (produced if necessary) at X and Y respectively.

Prove $XY = CD$

If ABCD is a parallelogram, then $AB \parallel CD$ , $BC \parallel AD$ , $AB = CD$ and $AD = BC$ .

In $\Delta ABX\angle BAX = \angle BXA$ because

$\angle BAX = \angle XAY$ (AX is the bisector $\angle A$ - given)

$\angle BXA = \angle XAY$ (since $BC\|AD$ - given), so $\angle BAX = \angle BXA$

Hence $\Delta ABX$ is an isosceles triangle, so $AB = BX$

In $\Delta BAY\angle ABY = \angle AYB$ because

$\angle ABY = \angle YBX$ (BY is the bisector $\angle B$ - given)

$\angle AYB = \angle XBY$ (since $BC\|AD$ - given), so $\angle ABY = \angle AYB$

Hence $\Delta ABY$ is an isosceles triangle, so $AB = AY$

Hence $BX = AY$

In ABXY $BX = AY$ and $BX\|AY$ - given, so ABXY is a parallelogram and so $AB = XY$ , but $AB = CD$ - given, so $XY = CD$ .