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.

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

1 comment:

  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

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