Question 1195016
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Let P be the intersection of AG and BH.<br>
Angles A and B in the parallelogram are supplementary.  Angle BAP measure is half the measure of angle A; Angle ABP is half the measure of angle B; so angle APB is a right angle.<br>
That makes triangles BPA and BPG congruent by Angle-Side-Angle; so BG is congruent to AB.<br>
A similar argument makes AH congruent to AB, making ABGH a rhombus.<br>