**Problem 7 of the Belarus Mathematical Olympiad 2000**

On the side AB of a triangle ABC with BC < AC < AB, points B_1 and C_2 are marked so that AC_2 = AC and BB_1 = BC. Points B_2 on side AC and C_1 on the extension of CB are marked so that CB_2 = CB and CC_1 = CA. Prove that the lines C_1C_2 and B_1B_2 are parallel.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien**

Observe that BB_2 and C_1A are parallel as given by the problem which makes ∠B_2BB_1 = ∠C_1AC_2. All we need to do now is to prove that the two triangles B_2BB_1 and C_1AC_2 are similar which makes ∠C_2C_1A = ∠B_1B_2B and C_1C_2 || B_1B_2.

Given ∠B_2BB_1 = ∠C_1AC_2 as mentioned, we only need to prove \frac {BB_2}{AC_1} = \frac {BB_1}{AC_2}

But since BB_2 || AC_1, we have \frac {BB_2}{AC_1} = \frac {CB}{CC_1}

The problem also gives CB = BB_1 and CC_1 = AC = AC_2; therefore, \frac {CB}{CC_1} = \frac {BB_1}{AC_2}

or \frac {BB_2}{AC_1} = \frac {BB_1}{AC_2} and the proof is done.