$D$ and $E$ are respectively the points on the sides $A B$ and $A C$ of a $Δ A B C$ such that $A B = 12 c m, A D = 8 c m, A E = 12 c m$ and $A C = 18 c m$ , show that $D E ∥ B C$ .

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Solution

$I n △ A B C w e h a v e, A B = 12 c m, A C = 18 c m, A D = 8 c m, a n d A E = 12 c m ∴ B D = A B - A D = (12 - 8) c m = 4 c m a n d C E = A C - A E = (18 - 12) c m = 6 c m N o w B y t a k i n g r a t i o s o f s i d e s o f t r i a n g l e \frac{A D}{B D} = \frac{8}{4} = \frac{2}{1} a n d \frac{A E}{C E} = \frac{12}{6} = \frac{2}{1} \Rightarrow \frac{A D}{B D} = \frac{A E}{C E} T h u s, D E d i v i d e s s i d e s A B a n d A C o f △ A B C i n t h e s a m e r a t i o . A c c o r d o n g t o c o n v e r s e o f b a s i c p r o p o r t i o n a l i t y t h e o r m i f a l i n e d i v i d e s a n y t w o s i d e s o f a t r i a n g l e i n t h e s a m e r a t i o, t h e n t h e l i n e m u s t b e p a r a l l e l t o t h e t h i r d s i d e . T h e r e f o r e, b y t h e c o n v e r s e o f b a s i c p r o p o r t i n a l i t y t h e o r e m, w e h a v e D E ∥ B C$

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D and E are respectively the points on the sides AB and AC of a ΔABC such that AB=12cm,AD=8cm,AE=12cm and AC=18cm, show that DE∥BC.

$D$ and $E$ are respectively the points on the sides $A B$ and $A C$ of a $Δ A B C$ such that $A B = 12 c m, A D = 8 c m, A E = 12 c m$ and $A C = 18 c m$ , show that $D E ∥ B C$ .