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In triangle $A B C, ∠ B A C = 90^{\circ},$ and $A D$ is its bisector. If $D E$ is drawn $⊥ A C$ , prove that $D E \times (A B + A C) = A B \times A C$ .

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Solution

$G i v e n : △ A B C i n w h i c h ∠ B A C = 90, A D i s t h e b i s e c t o r . A l s o D E ⊥ A C T o P r o v e : D E \times (A B + A C) = A B \times A C . P r o o f : I t i s g i v e n t h a t A D i s t h e b i s e c t o r o f ∠ A o f △ A B C ∴ \frac{A B}{A C} = \frac{B D}{D C} (B y A n g l e B i s e c t o r T h e o r m) \Rightarrow \frac{A B}{A C} + 1 = \frac{B D}{D C} + 1 [A d d i n g 1 o n b o t h s i d e s] \Rightarrow \frac{A B + A C}{A C} = \frac{B D + D C}{D C} \Rightarrow \frac{A B + A C}{A C} = \frac{B C}{D C} . . . . . . . . . (1) I n △ C D E a n d △ C B A, w e h a v e ∠ D C E = ∠ B C A [C o m m o n] ∠ D E C = ∠ B A C [E a c h e q u a l t o 90 S o, b y A A - c r i t e r i o n o f s i m i l a r i t y △ C D E \sim △ C B A \Rightarrow \frac{C D}{C B} = \frac{D E}{B A} (B y A n g l e B i s e c t o r t h e o r m) \Rightarrow \frac{A B}{D E} = \frac{B C}{D C} . . . . . (2) F r o m (1) a n d (2) w e h a v e \frac{A B + A C}{A C} = \frac{A B}{D E} \Rightarrow D E \times (A B + A C) = A B \times A C . H e n c e P r o v e d$

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In figure, $∠ B A C = 90^{o}$ , AD is its bisector. If DE $⊥$ AC, Prove that $D E \times (A B + A C) = A B \times A C$ .

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