## Saturday, August 4, 2018

### Geometry Problem 1373: Isosceles Triangle, Exterior Cevian, Inradius, Exradius, Altitude to the Base

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. Denote center of small circle as P and large circle as Q
Drop a perpendicular from P to BE,denote it as X and form a right-triangle BPX
From Q, drop perpendicular to BC,denote it as Y and form a right-trianlge BQY
We will prove that these two triangles are congruent and the result follows

Let the foot of perpendicular from P to DA be G and similarly from Q to DC be H
Similarly the foot of Perpendicular from P to DB be U and Q to DB be V
Let AB=BC=a , AE=EC=b, GA=x, CH=y
We know BV=BY=GE=PX=x+b ---------------(1)

let m(BPA)=@ and m(BAC)=\$
=>m(XBP)=90-\$+@ and m(BPX)=\$-@ ------------(2)
Similarly m(QBY)=\$-@ and m(BQY)=90-\$+@-------------(3)

From (1),(2) and (3), BPX and QBY are congruent and BX=QY=r2
Since XE=r1
=>h=BX+XE=r2+r1 Q.E.D

1. Correction :

2. How do we know BY=GE?

3. Let BA=BC=a, UB=a1 and BV=a2 and r1 touch AB at Z
then UB=BZ=a1 => AG=a-a1, ||ly BV=BY=a2 => CH=a-a2 -------(1)
But UV=GH (since DV=DH)
=> a1+a2=GA+AE+EC+CH
=> a1+a2=a-a1+b+b+a-a2
=> a1+a2 = a+b
=> GH = a+b--------(2)

Now GE= GH-EH
=> GE=(a+b)-(b+a-a2)
=> GE=a2=BY

2. Is the reverse true? I mean, if the height of the triangle is the sum of radii of the circles; is it always an isosceles triangle? I couldn't prove this.