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

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

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## Wednesday, August 8, 2018

### Geometry Problem 1375: Isosceles Triangle, Interior Cevian, Exradius, Excircle, Altitude to the Base

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Let the smaller circle be P and the larger Q.

ReplyDeleteLet 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

https://photos.app.goo.gl/qkDfNLjerBjioR1u9

ReplyDeleteDenote 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

Let center of small circle be P and the larger be Q

ReplyDeleteExtend 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