Criteria for kite

I have spotted that if in a quadrilateral $ABCD$ , if $\angle ABC = \angle CDA$ and $AC \perp BD$ , then quadrilateral $ABCD$ is a kite. This result is also used in APMO 2007 Q2.

I tried but couldn't prove it.

It will be a big help if someone can please show the proof.

#Geometry

Easy Math Editor

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Comments

I drew a picture of a kite and labled the points. I found out that it's true that AC is perpendicular to BD but I doubt that the measures of angles ABC and CDA are equal. If ABC and CDA have equal angle measures, then I think that the kite would turn into a rhombus.

A Former Brilliant Member - 2 years, 6 months ago

ABCD is a quadrilateral with angle BAC & angle BCD are equal and AC perpendicular to BD. How to prove that ABCD is kite?

SANTU PAUL - 2 years, 6 months ago

Well, I don’t really know how to prove that ABCD is a kite.

A Former Brilliant Member - 2 years, 6 months ago

So you can prove it a bit logically. You are given a $\triangle ABC$ and an altitude $BL$ . The locus of all points in plane such that $\angle CDA= \angle ABC$ is an arc of a circle with $BC$ as a chord. rite? This arc can intersect BL exactly only at this Pont. So it's $D$ .

Vishwash Kumar ΓΞΩ - 2 years, 5 months ago

So we'll start with the given condition that AC is perpendicular to BD. Now in order to prove that ABCD is a kite, we have to show that one of the perpendicular bisects the other and we'll use the fact ∠ABC = ∠CDA. Draw the two perpendiculars AC and BD such that they intersect at a point (and as arbitrarily as possible such that if you join the 4 end pts it does not look like a kite) Now connect AB and BC and note the angle ABC. Now take the lines AB and BC drawn and flip them across the page (or down depending which way you drew your perpendiculars) to form the points A'B'C', translate the point B' to the point D. This fulfills the criteria ∠ABC = ∠CDA. You should be able to see that the resulting figure formed falls short of being a quadrilateral and the only way that a quadrilateral can be formed in accordance to the two criteria above is if the point of intersection of the two perpendiculars - E - lies on the midpoint of AC and therefore BD bisects AC, and thus the resulting figure ABCD is a kite.

Zhang Xiaokang - 2 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$