## Wednesday, August 8, 2018

### Geometry Problem 1375: Isosceles Triangle, Interior Cevian, Exradius, Excircle, 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. Let the smaller circle be P and the larger Q.
Let PD meet BE at F.
Let BD meet circle P at U and circle Q at V. Let AC meet circle P at X and circle Q at Y.
Let AB=a=BC, AE=EC=d, AX=b and CY=c.

Now a+b = x+y ............(1)

Also DV = DY so, y+a-c = d-b-x+d+c
Therefore c-b = a-d ...(2) where we have used (1) above

Now EF = dh/(a+d) and so from similar triangles

b/r1=c/r2=h/(a+d)=(c-b)/(r2-r1)=(a-d)/(r2-r1) from (2) above

Hence h(r2-r1) = (a+d)(a-d) = a^2-d^2 = h^2

Therefore r2 - r1 = h

Sumith Peiris
Moratuwa
Sri Lanka

2. https://photos.app.goo.gl/qkDfNLjerBjioR1u9
Denote AB=BC= a
EC=EB=b
CF= x and AL=AH=y
We have BG=a-x and BH=BM=a+y
EL=b-y
So MG=2a-x+y and LF= 2b+x-y
Equate MG= LF we have b+x=a+y => EF=BH=BM=BN
We have ∠ (GBC) is an external angle of triangle DBC
And ∠ (BDC) is and external angle of triangle ABD
and BO2 and BO1 are angle bisectors of angle GBC and angle HBM
With angles manipulation we will have ∠ (MBO1) = ∠ (BNO2)
Triangle BMO1 congruent to BNO2 ( case SAS)
So NO2=MO1= r1 and r2-r1= NF=BE= h

3. Let center of small circle be P and the larger be Q
Extend BE to meet the perpendicular from P at X, join BP
Drop Perpendicular from Q to BC and denote it as Y, join BQ

Let m(ABE)=m(CBE)=@
and m(EBD)=\$
=> m(XBP)=(@+\$)/2 , m(BPX)=90-(@+\$)/2 -------------(1)
Similarly m(YBQ)=90-(@+\$)/2 and m(YQB)=90-(@+\$)/2 -------------(2)
Also it could be derived that BY=XP -----------------(3)

Hence Tr.BPX is congruent with QBY => BX=QY
=> h+r1=r2 Q.E.D

------------------------------------------------------------------------
To derive BY=XP

Let Perpendicular from Q to BC be F,from P to AC be L
Let the common tanget through B touch both smaller and larget circles at M and G
Similarly BA is tanget to P at H and let AB=BC=a,AE=EC=b, AH=AL=x and CF=CY=y

MG=MD+DG
=> MG=DL+DF -----------(1)

Also MG=GB+BM
=>MG=BY+BH
=>MG=a-y+a+x ----------------(2)

(1)=(2)
=>DL+DF=2a+x-y
=>b-x+b-y=2a+x-y
=>b-x=a-y
=>EL=BY
=>XP=BY