document.write( "Question 1195016: Given a parallelogram □ABCD, with AD > AB.
\n" ); document.write( "The bisector of ∠A intersects 𝐵𝐶 at G, and the bisector of ∠B intersects𝐴𝐷 at H.
\n" ); document.write( "Prove that □ABGH is a rhombus.
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Algebra.Com's Answer #827369 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "This is one way to draw out the diagram to supplement the answer @greenestamps posted
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\n" ); document.write( "The 90 degree angle can be found using the scratch work shown at the bottom of the solution page.
\n" ); document.write( "We must have AD be longer than AB, otherwise the points G and H wouldn't be possible.\r
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\n" ); document.write( "\n" ); document.write( "It then leads to this:
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\n" ); document.write( "Note that if x+y = 90 and y+z = 90, then we can say x = z.
\n" ); document.write( "This line of thinking helps determine the missing blue and red angles.\r
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\n" ); document.write( "\n" ); document.write( "This shows we have four congruent right triangles.
\n" ); document.write( "Each hypotenuse is the same, hence all four sides of this sub-figure are the same length.
\n" ); document.write( "This proves ABGH is a rhombus.\r
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\n" ); document.write( "\n" ); document.write( "-----------------------------\r
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\n" ); document.write( "\n" ); document.write( "Scratch Work:
\n" ); document.write( "Angle HAB + Angle ABG = 180
\n" ); document.write( "(Angle HAP + Angle PAB) + (Angle ABP + Angle PBG) = 180
\n" ); document.write( "(Angle PAB + Angle PAB) + (Angle ABP + Angle ABP) = 180
\n" ); document.write( "2*(Angle PAB) + 2*(Angle ABP) = 180
\n" ); document.write( "2*(Angle PAB + Angle ABP) = 180
\n" ); document.write( "Angle PAB + Angle ABP = 180/2
\n" ); document.write( "Angle PAB + Angle ABP = 90
\n" ); document.write( "Hence, triangle APB is a right triangle. This means all angles around point P are 90 degree angles as well.
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