SOLUTION: Given a parallelogram □ABCD, with AD > AB. The bisector of ∠A intersects 𝐵𝐶 at G, and the bisector of ∠B intersects𝐴𝐷 at H. Prove that □ABGH is a rhombus.

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Question 1195016: Given a parallelogram □ABCD, with AD > AB.
The bisector of ∠A intersects 𝐵𝐶 at G, and the bisector of ∠B intersects𝐴𝐷 at H.
Prove that □ABGH is a rhombus.
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website!

Let P be the intersection of AG and BH.

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.

That makes triangles BPA and BPG congruent by Angle-Side-Angle; so BG is congruent to AB.

A similar argument makes AH congruent to AB, making ABGH a rhombus.

Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

This is one way to draw out the diagram to supplement the answer @greenestamps posted

The 90 degree angle can be found using the scratch work shown at the bottom of the solution page.
We must have AD be longer than AB, otherwise the points G and H wouldn't be possible.

It then leads to this:

Note that if x+y = 90 and y+z = 90, then we can say x = z.
This line of thinking helps determine the missing blue and red angles.

This shows we have four congruent right triangles.
Each hypotenuse is the same, hence all four sides of this sub-figure are the same length.
This proves ABGH is a rhombus.

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Scratch Work:
Angle HAB + Angle ABG = 180
(Angle HAP + Angle PAB) + (Angle ABP + Angle PBG) = 180
(Angle PAB + Angle PAB) + (Angle ABP + Angle ABP) = 180
2*(Angle PAB) + 2*(Angle ABP) = 180
2*(Angle PAB + Angle ABP) = 180
Angle PAB + Angle ABP = 180/2
Angle PAB + Angle ABP = 90
Hence, triangle APB is a right triangle. This means all angles around point P are 90 degree angles as well.