## Sunday, August 5, 2018

### Geometry Problem 1374: Isosceles Triangle, Exterior Cevian, Incircle, Excircle, Tangency Points, Parallel Lines

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. Let the altitude BKL of Tr. ABC meet AC at L and the parallel to AC thro E at K.

Let BL = h, AB = BC = a, BF = a - b, CH = c, AL = LC = d

Let < FBE = < FKE = x, BFEK being cyclic
Let < KHL = y

As per my solution to Problem 1373, BK = h - r1 = r2 = GH
So BKHG is a parallelogram and so KH //BG

Now tan x = r1/(a-b) and tan y = r1/(c + d)

But since a - b = c + d again as per problem 1373, tan x = tan y
Hence x = y
So FKH must be collinear since EK // AH

So finally FKH // BG

Sumith Peiris
Moratuwa
Sri Lanka

2. Drop a Perpendicular from B to AC and meet at P
Drop perpendicular from E to BP and meet at Q
Join BE and let m(ABE)=m(EBF)=@ and m(BAC)=m(BCA)=\$ => m(GBC)=\$-@ -------------(1)
Connect QH and form the right tr. QPH
In Right tr.QPH, QP=EF (radius of small circle) ------(2)
PH=BF------------(3)
Hence QPH and EFB are congruent (SAS)
=>m(QHP)=@ ---------------(4)

Since m(EFB)=m(EQB)=90 => BFEQ are concylic
=>m(FBE)=m(FQE)=@ ---------------(5)

From (4) and (5), F,Q and H are collinear
Simple angle chasing in tr.BDA gives m(BDA)=m(FDH)=\$-2@
Hence, m(HFD)=180+@-\$------------------(5)
We know m(GBD)=180-m(GBC)=180-\$+@---------------(6)

From (5) and (6) BG//HF