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Let PQRS be a quadrilateral. If M and N are themid points of the sides PO and RS respectivelythen prove that PS+QR=2MN. |
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Answer» Answer: Let \OVERLINE{m}, \overline{n}, \overline{p}, \overline{Q}, \overline{r} m , n , p , q , r and \overline{s} s be position vectors of M,N,P,Q,RM,N,P,Q,R and SS respectively. MM and NN are midpoints of PQPQ and RSRS \therefore \overline {m} = \dfrac {\overline {p} + \overline {q}}{2}∴ m = 2 p + q
and \overline {n} = \dfrac {\overline {r} + \overline {s}}{2} n = 2 r + s
\overline {PS} + \overline {QR} = \overline {s} - \overline {p} + \overline {r} - \overline {q} PS + QR = s − p + r − q
= (\overline {r} + \overline {s}) - (\overline {p} + \overline {q})=( r + s )−( p + q ) = 2\overline {n} - 2\overline {m}=2 n −2 m
= 2(\overline {n} - \overline {m})=2( n − m ) = 2\overline {MN}=2 MN Step-by-step explanation: |
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