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Diagonals of a Rhombus Bisect Each-Other at Right Angles

ABCD is a rho...

Question

ABCD is a rhombus. Show that diagonal AC bisects $∠$ A as well as $∠$ C and diagonal BD bisects $∠$ B as well as $∠$ D.

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Solution

$\begin{matrix} [I n r h o m b u s a l l s i d e a r e e q u a l a n d d i a g o n o l s a r e d i f f e r e n t b u t i n t e r sec t a t 90^{0}] \Rightarrow A B = B C = C D = D A \Rightarrow A C ⊥ B D I n Δ A B C \Rightarrow A B = B C \Rightarrow∴ ∠ C A B = ∠ A C B \Rightarrow A O = O C ∴ ∠ A B O = ∠ C B O a n d s i m i l l a r l y i n Δ A D C \Rightarrow A D = D C \Rightarrow A O = O C ∴ c o r r e s p o n d i n g a n g l e a r e e q u a l \Rightarrow ∠ D A C = ∠ D C A a n d ∠ C D O = ∠ A D O H e n c e, A C b i sec t ∠ A a n d ∠ C a n d B D b i sec t s ∠ B a n d ∠ D \end{matrix}$

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A B C D

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In a rhombus ABCD show that diagonal AC bisects $∠ A$ as well as $∠ C$ and diagonal BD bisects $∠ B$ as well as $∠ D$ .

∠ A

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