In the adjoining figure- D- E- F are the midpoints of the sides BC- CA and AB of - ABC- If BE and DF intersect at X while CF and DE intersect at Y- prove that XY-1-4BC-

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In the adjoin...

Question

In the adjoining figure, $D, E, F$ are the midpoints of the sides $B C, C A a n d A B$ of $Δ A B C$ . If $B E a n d D F$ intersect at $X$ while $C F a n d D E$ intersect at $Y$ , prove that $X Y = \frac{1}{4} B C$ .

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Solution

$G i v e n : △ A B C w h e r e D, E, F a r e t h e m i d p o i n t s o f t h e s i d e s B C, C A a n d A B o f △ A B C . T o P r o v e : X Y = \frac{1}{4} B C P r o o f : I n △ A B C, F a n d E a r e t h e m i d p o i n t s o f A B a n d A C r e s p e c t i v e l y ∴ F E ∥ B C a n d F E = \frac{1}{2} B C = B D (T h e l i n e j o i n i n g t h e m i d - p o i n t s o f t w o s i d e s o f a t r i a n g l e i s p a r a l l e l t o t h e t h i r d s i d e a n d e q u a l t o h a l f t h e l e n g t h o f t h e t h i r d s i d e) ∴ F E ∥ B D a n d F E = B D . H e n c e B D E F i s a p a r a l l e l o g r a m w h o s e d i a g o n a l s B E a n d D F i n t e r s e c t e a c h o t h e r a t X) ∴ X i s t h e m i d p o i n t o f D F . S i m i l a r l y, Y i s t h e m i d p o i n t o f D E . T h u s, i n △ D E F, X a n d Y a r e t h e m i d p o i n t s o f D F a n d D E r e s p e c t i v e l y . S o, X Y ∥ F E a n d X Y = d \frac{1}{2} F E (M i d p o i n t T h e o r m) = \frac{1}{2} \times \frac{1}{2} B C = \frac{1}{4} B C H e n c e P r o v e d$

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