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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!

 #1
avatar+99592 
+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 

 

 

 

cool cool cool

 Mar 4, 2018
edited by CPhill  Mar 4, 2018
 #3
avatar+1432 
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Thank you so much!

 #4
avatar+68 
+1

Just for anyone who doesn't want to draw the pic. ;)

CoopTheDupe  Mar 6, 2018

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