Prove a convex quadrilateral with perpendicular diagonals and one pair of congruent, non-consecutive angles is a kite.

Question

I strongly suspect the conditions in the title are sufficient for a kite, but I am unable to give a proof, direct or otherwise. I've attempted to proceed directly with triangle congruence strategies and trigonometry, by contraction assuming incongruent edges or the diagonal between congruent angles not bisected by the other diagonal, and even gotten nowhere with a coordinate geometry attempt.

I appreciate any and all help! I feel like I've gotten closest by assuming the latter contradiction and extending the figure into a rhombus, but that still doesn't quite get me there.

Answer 1

You can do this with only beginning triangle geometry, like Book I of Euclid's Elements, no need for circles or symmetry (though those are quick routes if you have those theorems handy).

We are given the convex quadrilateral at left with diagonals intersecting at right angles, and with $\angle{A} = \angle{C}$. I say $\triangle{ABD} \cong \triangle{CBD}$. For if not, then cut $HC'$ equal to $AH$. Then by SAS $\triangle{AHB} \cong \triangle{C'HB}$ and likewise $\triangle{AHD} \cong \triangle{C'HD}$. By addition of these like triangles, $\triangle{ABD} \cong \triangle{C'BD}$ and so $\angle{BAD} \cong \angle{BC'D}$. But we also have $\angle{A} = \angle{C}$, yet $\angle{A}=\angle{BC'D}$, so $\angle{C}=\angle{BC'D}$ the lesser to the greater, which is absurd. Therefore $AH=CH$ and $\triangle{ABD} \cong \triangle{CBD}$.

The key here is you get to assume something extra (that is false) which lets you solve the problem conclusively.

Answer 2

Let the angles at $A$ and $C$ be congruent, and diagonal $BD$ to intersect $AC$ at $H$. Point $A$ lies on an arc of circle $BAD$, and the locus of points $C$ such that $\angle BCD\cong\angle BAD$ is the reflection of that arc about $BD$. It follows that $AH\cong CH$ and so on.

Answer 3

Given that our diagonals are perpendicular, the fact that our opposite angles are equal gives the fact that the diagonal different from the segment relying the points having congruent angles is the perpendicular bisector of that segment. Hence, the two other points of our quadrilateral wherever they may be are on the perpendicular bisector of our segment and hence equidistant from its extremities. We have now a quadrilateral whose sides are 2 by 2 equidistant and adjacent, and so our quadrilateral is a kite.