Please solve 13th sum

Please solve 13th sum 13. Show that the line segments joining the mid points of two sides ofa triangle is parallel to the third side and half the length of that of the third

D e a r s t u d e n t A s y o u h a v e u p l o a d e d t h e q u e s t i o n u n d e r v e c t o r s . I a m p r o v i n g i t u \sin g v e c t o r s .

C h o o s e o n e o f t h e v e r t e x a s o r i g i o n O . L e t o t h e r v e c t o r s b e A (\vec{a}) a n d B (\vec{b}) . L e t C (\vec{c}) a n d D (\vec{d}) b e m i d p o i n t s o f s i d e O A a n d O B r e s p e c i t v e l y . \vec{A B} = \vec{b} - \vec{a} S i n c e C a n d D a r e m i d p o i n t o f O A a n d O B r e s p e c t i v e l y, u \sin g m i d p o i n t f o r m u l a w e h a v e \vec{c} = \frac{\vec{0} + \vec{a}}{2} = \frac{\vec{a}}{2} \vec{d} = \frac{\vec{0} + \vec{b}}{2} = \frac{\vec{b}}{2} \vec{C D} = \vec{d} - \vec{c} = \frac{\vec{b} - \vec{a}}{2} = \frac{1}{2} (\vec{b} - \vec{a}) \vec{C D} = \frac{1}{2} \vec{A B} \Rightarrow |\vec{C D}| = |\frac{1}{2} \vec{A B}| a n d \vec{C D} | | \frac{1}{2} \vec{A B} \Rightarrow |\vec{C D}| = \frac{1}{2} |\vec{A B}| a n d \vec{C D} | | \vec{A B} H e n c e, C D = \frac{1}{2} A B a n d C D | | A B H e n c e, w e h a v e p r o v e d t h e r e s u l t .