A circle is drawn that intersects all three sides of triangle PQR as shown below. Prove that if AB = CD = EF, then the center of the circle is the incenter of triangle PQR.

Please include a diagram if you introduce any new information into your solution.

Thanks so much!

AnonymousConfusedGuy
Mar 4, 2018

#1**+3 **

Find the center of the circle.....call this " O"

From O.....draw bisector OG to chord AB and bisector OH to chord EF

But, by Euclid, a radial line bisecting a chord does so at right angles

And since AB = EF = CD.....then equal chords are equal distances from the center of a circle

Now....connect O and P

So.....OG = OH, OP = OP and angles OGP and OHP are right

So, by HL, have two cogruent right triangles formed, ΔOGP and ΔOHP

But angle GPO = angle HPO....so angle P is bisected

By similar reasoning, we can draw bisector OI to chord CD and also connect QO

This will form two congruent right triangles - Δ OGQ and ΔOIQ

And angle GQO = angle IQO, so Q is bisected

And since PO and QO are angle bisectors of triangle QPR that meet at O....then ,by definition, the center of the circle is the incenter of triangle PQR

CPhill
Mar 4, 2018