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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below for more details.
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. HDSince HA=HC => p1-p2= HDWe have DE-DF= ( p1-BA)-(p2-BC)= p1-p2 since BA=BCSo 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. Let EH = x, DF = y and FC = zNow AE = AP = b/2 - x where AC = bBP = BQ = a+x- b/2 where AB = BC = aFurther DF = DR and CS = CF and DE = DQSo further since BR = BS, (a+x-b/2) + (b/2 -y-z + x) - y = a - zSimplifying x = y so EH = DFSumith PeirisMoratuwaSri Lanka
It can further be seen that1) < HBO1 = < DBO22) DO1 and DO2 are perpendicular to each other3) AO1 and CO2 meet on BH
I extended the problem as follows:http://www.fastpic.jp/images.php?file=4938134885.gif
We already had this problem and solution. see Problem 560 in the link below.http://www.gogeometry.com/problem/p560_triangle_cevian_incircles_tangent_congruence.htmPeter Tran
Problem 349 is another extended problemhttp://gogeometry.com/problem/p349_triangle_cevian_incircles_common_tangent.htmThanks Antonio