document.write( "Question 1195016: Given a parallelogram □ABCD, with AD > AB.
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document.write( "The bisector of ∠A intersects 𝐵𝐶 at G, and the bisector of ∠B intersects𝐴𝐷 at H.
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document.write( "Prove that □ABGH is a rhombus. \n" );
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Algebra.Com's Answer #827364 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Let P be the intersection of AG and BH. \n" ); document.write( "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. \n" ); document.write( "That makes triangles BPA and BPG congruent by Angle-Side-Angle; so BG is congruent to AB. \n" ); document.write( "A similar argument makes AH congruent to AB, making ABGH a rhombus. \n" ); document.write( " \n" ); document.write( " |