In the adjoining figure, $A B C D$ is a trapezium in which $A B ∥ D C a n d A D = B C$ . If $P, Q, R, S$ be respectively the midpoints of $B A, B D a n d C D, C A$ then show that $P Q R S$ is a rhombus.

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Solution

$G i v e n : T r a p e z i u m A B C D w i t h A B ∥ D C a n d A D = B C . P, Q, R, S a r e r e s p e c t i v e l y t h e m i d p o i n t s o f B A, B D, C D a n d C A . T o P r o v e : P Q R S i s a r h o m b u s P r o o f : I n △ B D C, Q a n d R a r e t h e m i d p o i n t s o f B D a n d C D r e s p e c t i v e l y ∴ Q R ∥ B C a n d Q R = \frac{1}{2} B C [A c c o r d i n g t o M i d p o i n t T h e o r e m] I n △ A B C, P a n d S a r e m i d p o i n t s o f B A a n d A C r e s p e c t i v e l y ∴ P S ∥ B C a n d P S = \frac{1}{2} B C (A c c o r d i n g t o M i d p o i n t T h e o r e m) T h u s, P S ∥ Q R a n d P S = Q R [E a c h e q u a l t o \frac{1}{2} B C] ∴ P Q R S i s a p a r a l l e l o g r a m . I n △ A C D, S a n d R a r e t h e m i d p o i n t s o f A C a n d C D r e s p e c t i v e l y . S R ∥ A D a n d S R = \frac{1}{2} A D = \frac{1}{2} B C [∵ A D = B C (G i v e n)] I n △ A B D, P a n d Q a r e t h e m i d p o i n t s o f A B a n d B D r e s p e c t i v e l y . P Q ∥ A D a n d P Q = \frac{1}{2} A D = \frac{1}{2} B C [∵ A D = B C (G i v e n)] ∴ P S = Q R = S R = P Q . H e n c e, P Q R S i s a r h o m b u s$ .

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