In a parallelogram $A B C D$ , the bisector of $∠ A$ also bisects $B C$ at $X$ . Prove that $A D = 2 A B$ .

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Solution

$\begin{matrix} L e t T h e b i sec t o r o f ∠ A b i sec t s t h e s i d e B C a t x . G i v e n; - A B C D i s p a r a l l e o g r a m ∴ A D ∥ B C N o w, A X i s t h e t r a n s v e r s a l ∴ ∠ 1 = ∠ 3 (a l t e r n a t e a n g l e s) - - - - (1) a n d ∠ 1 = ∠ 2 (A X i s t h e b i sec t o r o f ∠ A) - - (2) F r o m (1) a n d (2) w e g e t ∠ 1 = ∠ 3 N o w, I n Δ A B X ∠ 1 = ∠ 3 \Rightarrow A B = B X \Rightarrow 2 A B = 2 B X = B X + B X = B X + X C (X i s t h e m i d p o i n t s o f B C) \Rightarrow 2 A B = B C \Rightarrow 2 A B = A D (o p p o s i t e s i d e o f a p a r a l l e log r a m a r e e q u a l) ∴ A D = 2 A B p r o v e d . \end{matrix}$

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