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Question
In a parallelogram ABCD, M is the mid-point AC. X and Y are the points on AB and DC respectively such that AX = CY. Prove that:
(i) Triangle AXM is congruent to triangle CYM, and
(ii) XMY is a straight line.
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Solution
(i) Join XM and MY.
In ΔAXM and ΔCYm
AM = MC ...(given)
AX = CY ...(given)
∠XAM = ∠YCM ...(alternate angles)
Therefore, ΔAXM ≅ ΔCYM.
(ii) ∠AMX + ∠AMY = 180° ...(linear pair of angle = 180°)
THerefore, XMY is a straight line.
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