In the given figure, $P A T$ is a tangent to the circumcircle of a $Δ A B C$ at the vertex $A$ . A line parallel to $P A T$ intersects the sides $A B$ and $A C$ at the points $D a n d E$ respectively.
Prove that
$Δ A B C \sim Δ A E D$
$A B \cdot A D = A C \cdot A E$

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Solution

$G i v e n : P A T i s a t a n g e n t t o t h e c i r c u m c i r c l e o f a △ A B C a t t h e v e r t e x A . A l i n e p a r a l l e l t o P A T i n t e r s e c t s t h e s i d e s A B a n d A C a t t h e p o i n t s D a n d E r e s p e c t i v e l y . T o P r o o f : △ A B C \sim A E D A B A D = A C A E . P r o o f : ∠ T A C = ∠ T A E = ∠ A B C . . . . (1) [∵ a n g l e s i n t h e a l t e r n a t e s e g m e n t a r e e q u a l] S i m i l a r l y, ∠ P A B = ∠ P A D = ∠ A C B . . . . (2) [∵ A n g l e s i n t h e a l t e r n a t e s e g m e n t a r e e q u a l] . A s P T ∥ D E A l s o ∠ T A E = A E D (a l t e r n a t e a n g l e s) . . . . . (3) ∠ P A D = ∠ A D E [A l t e r n a t e a n g l e s] . . . . . (4) F r o m (1) t o (4), w e g e t ∠ A B C = ∠ A E D ∠ A C B = ∠ A D E ∴ △ A B C \sim ∠ A E D [B y A A s i m i l a r i t y] S i n c e c o r r e s p o n d i n g s i d e s o f s i m i l a r t r i a n g l e s a r e p r o p o r t i o n a l . \Rightarrow \frac{A B}{A E} = \frac{A C}{A D} \Rightarrow A B \times A D = A C \times A E H e n c e, t h e r e q u i r e d r e s u l t s a r e p r o v e d .$

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In the given figure, PAT is a tangent to the circumcircle of a ΔABC at the vertex A. A line parallel to PAT intersects the sides AB and AC at the points DandE respectively.Prove thatΔABC∼ΔAEDAB⋅AD=AC⋅AE

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