## Problem 4 from the IMO 2009

Here is Problem 4 from the IMO 2009:

Let ABC be a triangle with AB = AC. The angle bisectors of ∠CAB and ∠ABC meet the sides BC and CA at D and E, respectively. Let K be the incentre of triangle ADC. Suppose that |

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny### Solution by Vo Duc Dien

(Saturday July 18, 2009)

Dedicated to my teacher Nguyen Hoi

Extend CK to meet BE at H. H is the center of the incircle of triangle ABC.

Since CH bisects ∠ACB, the two right triangles HDC and HMC are equal. Therefore

There are two possibilities: either M and E coincide or they are different. In the former case, BM is both the median and the altitude from B, implying *BC = AB.* Since also, *AB = AC,* all three sides in ΔABC are equal and the triangle is equilateral. Thus

If M and E are distinct, we recall that ∠HEK = ∠BEK = 45°. Thus the quadrilateral HMEK is cyclic and, since

Further, ∠BHC = 180° - ∠KHE = 135°, 2b = 45°.

In triangle ABC we have ∠CAB = 180° - 4b = 90° which is the second possible value for ∠CAB.

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny61210476 |