SOLUTION: Prove by contradiction that the diagonals of a kite intersect at right angles? My proof is very wordy,can you help me be more precise. Thank you

Question 1206126: Prove by contradiction that the diagonals of a kite intersect at right angles? My proof is very wordy,can you help me be more precise. Thank you
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Consider kite ABCD.
By definition of a kite, adjacent sides are congruent so and where . Not all 4 sides are congruent, otherwise we would have a rhombus.

One key property of a kite is that one diagonal is cut in half by the other. Both diagonals are not bisected (otherwise we would have a parallelogram). Let's consider diagonal AC is bisected to get AE = EC where E is the intersection of the diagonals.

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Proof by Contradiction:
Assume the diagonals do not intersect at right angles.
It means neither angle AEB nor BEC is 90 degrees and furthermore

So either,
angle AEB > angle BEC
or
angle AEB < angle BEC

If angle AEB > angle BEC, then AB > BC due to the Hinge Theorem (see links below).
But this contradicts AB = BC.

If angle AEB < angle BEC, then AB < BC due to the Hinge Theorem which also contradicts AB = BC

Either contradiction leads us to conclude that the opposite of the assumption above must be the case. Therefore, we conclude that the diagonals of a kite are perpendicular.

Further reading about the Hinge Theorem
http://ceemrr.com/Geometry1/HingeTheorem/HingeTheorem_print.html
and
https://mathbitsnotebook.com/Geometry/SegmentsAnglesTriangles/SATHinge.html

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