$ABC$ is triangle. $l_1$- line passes through $A$ and parallel to $BC$, $l_2$ - line passes through $C$ and parallel to $AB$. Bisector of $\angle B$ intersect $l_1$ and $l_2$ at $X,Y$. $XY=AC$. What value can take $\angle A- \angle C$ ?
Problem
Source: St Petersburg Olympiad 2013, Grade 9, P3
Tags: geometry
RagvaloD
16.10.2017 14:35
Let $\angle A \leq \angle C$ Let $Z=AX\cup BY$. $\angle BCZ=\angle A+\angle C=180-\angle B$ and $\angle YBC=\frac{\angle B}{2}$ so $\angle BYC=\frac{\angle B}{2},BC=CY$ Same way we can prove, that $AX=AB, \angle YXZ=\frac{\angle B}{2}$ So $\triangle XYZ$ has angles $\frac{\angle B}{2},\frac{\angle B}{2},180-\angle B$ and sides equal $AC,BC-AB,BC-AB$ Let $BY$ intersect circumcircle of $ABC$ at $D$. Then $\triangle ADC =\triangle XZY$ Let $E$ - point on $BA$ such that $BE=BC \to AE=BC-AB=XZ=AD$ Let $F$ - point of intersection of circumcircle of $ABC$ and $CE$. $\angle BEC=90-\frac{\angle B}{2},\angle AFC=180-\angle B, \angle AFE=\angle B, \angle EAF=180-\angle B-(90-\frac{\angle B}{2})=90-\frac{\angle B}{2}=\angle AEF$, so $AF=EF$ $\angle AFD=\angle AFC+\angle CFD=180-\angle B+ \frac{\angle B}{2}=180-\frac{\angle B}{2}$ $\angle EFD=360-\angle AFD-\angle AFE =360-180+\frac{\angle B}{2}-\angle B=180-\frac{\angle B}{2}=\angle AFD$ So $\triangle EFD=\triangle AFD \to ED=AD=AE \to \angle BAD=120$ $120=\angle BAD=\angle A +\frac{\angle B}{2}=\angle A+\frac{180-\angle A-\angle C}{2}=\frac{180+\angle A-\angle C}{2} \to \angle A-\angle C=60$
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