Question
In the adjoining figure, ABCD is a parallelogram.
Line segments AX and CY bisect $\angle A$ and $\angle C$ respectively.
Prove that :
(i) $\triangle ADX \cong \triangle CBY$
(ii) $AX = CY$
(iii) $AX \| CY$
(iv) AYCX is a parallelogram.
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Answer

[Hint.
(i) $\angle D=\angle B, \angle D A X=\angle B C Y\left(\frac{1}{2} \angle A=\frac{1}{2} \angle C\right), A D=B C$.
(ii) $\angle 1=\angle 2\left(\frac{1}{2} \angle A=\frac{1}{2} \angle C\right)$, and $\angle 2=\angle 3$ (alt. int. $\left.\angle s \right) \Rightarrow \angle 1=\angle 3$.]

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