In $Δ A B C$ , the bisector of $∠ A$ intersects the base $B C$ at the point $D$ . Prove that $A B \times A C = B D \times D C + A D^{2}$ .

Open in App

Solution

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

Suggest Corrections

Similar questions

∠ C A F

Δ A B C

\frac{B A}{A C} = \frac{B D}{D C}

2 A B^{2} = 2 A C^{2} + B C^{2}

A D

∠ A

Δ A B C

A D

B C

D

\frac{B D}{C D} = \frac{A B}{A C}

∠

Join BYJU'S Learning Program