Question

Prove that the straight line joining the mid-points of an isosceles trapezium in order, form a rhombus.

Answer 1

Answer:

From the figure ∠DAB = ∠ABC since ABCD is isosceles trapezium ΔDAB ≅ ΔCAB (By SAS congruence axiom) Hence AC = BD In ΔDAB, P and S are midpoints of AB and AD respectively Hence SP||DB and Similarly, in ΔDCB QR||DB and Hence pair of opposite sides are equal and parallel. Hence it is a parallelogram Now consider, ΔABC P and Q are midpoints of AB and BC respectively, Hence PQ||AC and ∴PQ = SP Since a pair of adjacent sides are equal, PQRS is a rhombus

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