Thursday, September 24, 2015

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

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

Click the diagram below for more details.

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

2. 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.

Let 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

3. It can further be seen that
1) < HBO1 = < DBO2
2) DO1 and DO2 are perpendicular to each other
3) AO1 and CO2 meet on BH

4. I extended the problem as follows:
http://www.fastpic.jp/images.php?file=4938134885.gif

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

Peter Tran

2. Problem 349 is another extended problem

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

Thanks
Antonio

5. https://photos.app.goo.gl/oecijsFEWz6zFSFw9