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in Olympiad by (65.1k points)

Let ABCD be a convex quadrilateral P, Q, R, S be the mid points of AB, BC, CD, DA respectively, such that triangles AQR and CSP are equilateral. Prove that ABCD is a Rhombus. Determine its angles.

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Explanation:

We have QR = BD/2 = PS. Since AQR and CSP are both equilateral and QR = PS, They must be congruent triangles. This implies that AQ = QR = RA = CS = SP = PC. Also CEF=60°=RQA. (Fig. 1)

Hence CS is parallel to QA. Now CS = QA implies that CSQA is a parallelogram. In particular SA is parallel to CQ and SA = CQ. This shows that AD is parallel to BC and AD = BC. Hence ABCD is a parallelogram. Let the diagonal AC and BD bisect each other at W. Then DW = DB/2 = QR = CS = AR. Thus in triangle ADC, the medians AR, DW, CS are all equal. Thus ADC is equilateral. This implies ABCD is Rhombus. Moreover the angle are 60°and 120°.

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