Question

Let $ABC$ be a triangle with circum-circle $\Gamma$ . Let $M$ be a point in the interior of triangle $ABC$ which is also on the bisector of $\angle A.$ Let $AM,BM,CM$ meet $\Gamma$ in $A_1,B_1,C_1,$ respectively. Suppose $P$ is the point of intersection of $A_1{C}_1$ with $AB$, and $Q$ is the point of intersection of $A_1{B}_1$ with $AC.$ Prove that $PQ$ is parallel to $BC.$

Answer 1

Let $A=2\alpha$ .
Then $\angle A_{1}AC=\angle BA{A}_1=\alpha$ .
$\therefore \angle A_1{B}_{1}C=\alpha =\angle B{B}_1{A}_1=\angle A_1{C}_{1}C=\angle B{C}_1{A}_1$ .
We also have $\angle B_{1}CQ=\angle A{A}_1{B}_1=\beta$ , say.
It follows that triangles $M{A}_1{B}_1$ and $QC{B}_1$ are similar and
hence $\frac{QC}{M{A}_1}=\frac{B_{1}C}{B_1{A}_1}$ .
Similarly, triangles ACM and $C_1{A}_{1}M$ are similar and
we get $\frac{AC}{ACM}=\frac{C_1{A}_1}{C_{1}M}$ .
Using the point P, we get similar ratios : $\frac{PB}{M{A}_1}=\frac{C_{1}B}{A_1{C}_1}=\frac{AB}{AM}=\frac{A_1{B}_1}{M{B}_1}$
$\therefore \frac{QC}{PB}=\frac{A_1{C}_1.B_{1}C}{C_{1}B.B_1{A}_1}$ ,
and
$\frac{AC}{AB}=\frac{M{B}_1.C_1{A}_1}{A_1{B}_1.C_{1}M}$ .
$=\frac{M{B}_1{C}_1{A}_1}{C_{1}M{A}_1{B}_1}=\frac{M{B}_1{C}_{1}B.QC}{C_{1}MPB.B_{1}C}$
However, triangles $C_{1}BM$ and $B_{1}CM$ are similar, which gives
$\frac{B_{1}C}{C_{1}B}=\frac{M{B}_1}{M{C}_1}$
Putting this in the last expression, we get
$\frac{AC}{AB}=\frac{QC}{PB}$
We conclude that PQ is parallel to BC.