Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Saturday, August 4, 2018

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

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Denote center of small circle as P and large circle as Q

ReplyDeleteDrop 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

Correction :

Deletelet m(ABP)=@ instead of m(BPA)=@

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.

ReplyDelete