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

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

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## Thursday, September 24, 2015

### Geometry Problem 1149: Isosceles Triangle, Altitude, Cevian, Incircles, Tangent, Congruence

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Let 2.p1 and 2.p2 are perimeters of triangles ABD and BDC.

ReplyDeleteSince ABC is isosceles triangle => 2.p1- 2.p2= DA-DC= HA+HD-(HC-HD)= 2. HD

Since HA=HC => p1-p2= HD

We have DE-DF= ( p1-BA)-(p2-BC)= p1-p2 since BA=BC

So DE-DF= p1-p2= HD =HD+HE-DF = > HE=DF

Let the other 2 tangential points of O1 be P for AB and Q for BD. Similarly for O2, R for BD and S for AC.

ReplyDeleteLet EH = x, DF = y and FC = z

Now AE = AP = b/2 - x where AC = b

BP = BQ = a+x- b/2 where AB = BC = a

Further DF = DR and CS = CF and DE = DQ

So further since BR = BS,

(a+x-b/2) + (b/2 -y-z + x) - y = a - z

Simplifying x = y so EH = DF

Sumith Peiris

Moratuwa

Sri Lanka

It can further be seen that

ReplyDelete1) < HBO1 = < DBO2

2) DO1 and DO2 are perpendicular to each other

3) AO1 and CO2 meet on BH

I extended the problem as follows:

ReplyDeletehttp://www.fastpic.jp/images.php?file=4938134885.gif

We already had this problem and solution. see Problem 560 in the link below.

Deletehttp://www.gogeometry.com/problem/p560_triangle_cevian_incircles_tangent_congruence.htm

Peter Tran

Problem 349 is another extended problem

Deletehttp://gogeometry.com/problem/p349_triangle_cevian_incircles_common_tangent.htm

Thanks

Antonio