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Sine Rule

State the con...

Question

State the converse of Pythagoras' Theorem and prove it.

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Solution

Statement : If a right triangle, the square of the hypotenuse is equal to the sum of the square of other two sides.
$\begin{matrix} G i v e n Δ A B C r i g h t a n g l e a t B T o p r o v e - A C^{2} = A B^{2} + B C^{2} C o n s t r u c t i o n : D r a w B D ⊥ A C o o f : sin c e B D ⊥ A C B y u sin g t h e o r e m - i f a p e r p e n d i c u l a r i s d r a w n f r o m t h e v e r t e x o f t h e r i g h t a n g l e o f t h e a r i g h t t r i a n g l e t o t h e h y p o t e n u s e t h e n t r i a n e o n b o t h s i d e o f t h e p e r p e n d i c u l a r a r e s i m i l a r t o w h o l e t r i a n g l e a n d t o e a c h o t h e r . Δ A B D \sim Δ A B C sin c e, s i d e s o f s i m i l a r t r i a n g l e s a r e i n t h e s a m e r a t i o \Rightarrow \frac{A D}{A B} = \frac{A B}{A C} \Rightarrow A D . A C = A B^{2} - - - - (1) s i m i l a r l y Δ B D C \sim Δ A B C sin c e, s i d e s o f s i m i l a r t r i a n g l e s a r e i n t h e s a m e r a t i o \Rightarrow \frac{C D}{B C} = \frac{A B}{A C} \Rightarrow C D . A C = B C^{2} - - - - (2) N o w, a d d i n g (1 + 2) w e g e t A D . A C + C D . A C = A B^{2} + B C^{2} A C (A D + C D) = A B^{2} + B C^{2} A C \times A C = A B^{2} + B C^{2} ∴ A B^{2} + B C^{2} = A C^{2} p r o v e d . \end{matrix}$

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