Any point $D$ is taken in the base of a triangle $A B C$ and $A D$ is produced to $E$ , making $D E = A D$ . Show that $ar (Δ B C E) = ar (Δ A B C)$

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Solution

$\begin{matrix} T o p r o v e : A r e a o f Δ B C E = a r e a o f Δ A B C p r o o f : I n Δ A B E, A D = D E (g i v e n) ∴ B D i s a m e d i a n o f Δ A B E \Rightarrow a r (Δ A B D) = a r (Δ B E D) . . . . . . . (1) m e d i a n d i v i d e s a t r i a n g l e i n t o t w o t r i a n g l e s o f e q u a l a r e a . S i m i l a r l y, I n Δ A C E C D i s m e d i a n o f Δ C E D . . . . . . (2) A d d i n g (1) a n d (2), a r (Δ A B D) a r (Δ A C D) = a r (Δ B E D) + a r (Δ C E D) o r a r (Δ A B C) = a r (Δ B C E) H e n c e, a r e a o f Δ B C E = a r e a o f Δ A B C . \end{matrix}$

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