In a right-angled $Δ A B C$ a circle with side $A B$ as diameter is drawn to intersect the hypotenuse $A C$ in $L$ . Prove that the tangent to the circle at $L$ bisects the side $B C$ .

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Solution

$G i v e n : I n a r i g h t - a n g l e d △ A B C a c i r c l e w i t h s i d e A B a s d i a m e t e r i s d r a w n t o i n t e r s e c t t h e h y p o t e n u s e A C i n L T o P r o o f : T h e t a n g e n t t o t h e c i r c l e a t L b i s e c t s t h e s i d e B C . P r o o f : L e t t h e t a n g e n t a t L m e e t B C i n M . J o i n B L . F i r s t l y, ∠ A L B = 90 [A n g l e i n a s e m i c i r c l e] B L C = 90 S i n c e t h e l e n g t h s o f t a n g e n t s f r o m a n e x t e r n a l p o i n t a r e e q u a l . ∴ M B = M L N o w L M i s a t a n g e n t t o t h e c i r c l e a t L a n d L B i s a c h o r d t h r o u g h t h e p o i n t o f c o n t a c t L ∴ 3 = 2 F u r t h e r, ∠ M L C = ∠ B L C ∠ 3 = 903 a n d ∠ L C M = ∠ A C B = 902 = 903 [∵ ∠ 2 = ∠ 3] \Rightarrow M L C = L C M ∴ M L = M C T h u s B M = L M = M C . H e n c e, M i s t h e m i d - p o i n t o f B C . P r o v e d$

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