Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it Bulgaria

Problem:
Let $A_{1}$ , $B_{1}$ and $C_{1}$ be respectively the midpoints of the sides $BC$ , $CA$ and $AB$ of $\triangle ABC$ with centroid $M$ . The line through $A_{1}$ and parallel to $BB_{1}$ meets the line $B_{1}C_{1}$ at a point $D$ . Prove that if the points $A$ , $B_{1}$ , $M$ and $C_{1}$ are concyclic, then $\text{ADA 1 = CAB}$ .

Solution

Solution:
Since $A_{1}D \parallel MB_{1}$ and $A$ , $B_{1}$ , $M$ , $C_{1} \in k$ , it follows that
$\text{AA 1 D = AMB 1 = AC 1 B 1 = ABC.}$
Using that $B_{1}C_{1} \parallel BA_{1}$ and $A_{1}D \parallel BB_{1}$ we conclude that $BA_{1}DB_{1}$ is a parallelogram. Hence $BA_{1} = B_{1}D = B_{1}C_{1}$ . On the other hand, $AB_{1} = B_{1}C$ and therefore $AC_{1}CD$ is a parallelogram. In particular, $AD \parallel CC_{1}$ and then
$\text{DAA 1 = CMA 1 = AMC 1 = AB 1 C 1 = ACB.}$

Therefore
$\text{ADA 1 = 180 - DAA 1 - DA 1 A = 180 - ABC - ACB = CAB .}$

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