  Question

ABCD is a quadrilateral in which P, Q, R, and S are mid-points of the lines AB, BC, CD and DA ( see in fig ). AC is a diagonal. Show that:(i) SR || AC and $$SR =\frac{1}{2} AC$$(ii) PQ= SR(iii) PQRS is a parallelogram. Solution

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.$$(i)$$ In $$\bigtriangleup DAC$$ , S is the mid point of DA and R is the mid point of DC. Therefore, $$SR\parallel AC$$ and $$SR=\frac{1}{2}AC$$.By mid-point theorem.$$(ii)$$ In $$\bigtriangleup BAC$$ , P is the mid point of AB and Q is the mid point of BC. Therefore, $$PQ\parallel AC$$ and $$PQ=\frac{1}{2}AC$$.By mid-point theorem. But from $$(i)$$ $$SR=\frac{1}{2}AC$$ therefore $$PQ=SR$$$$(iii)$$  $$PQ\parallel AC$$ & $$SR\parallel AC$$ therefore  $$PQ\parallel SR$$ and $$PQ=SR$$. Hence, a quadrilateral with opposite sides equal and paralle is a parallelogram. Therefore PQRS is a parallelogram.Mathematics

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