Two non intersecting circles with centers $O_1$ and $O_2$ are tangent to line $s$ at points $A_1$ and $A_2$, respectively, and lying on the same side of this line. Line $O_1O_2$ intersects the first circle at $B_1$ and the second at $B_2$. Prove that the lines $A_1B_1$ and $A_2B_2$ are perpendicular to each other.
Problem
Source:
Tags: geometry, circles, perpendicular
AwesomeLife_Math
25.03.2020 20:26
Solution?
First label the intersection of the two lines $A_1B_1$ and $A_2B_2$ as $P$ and $\angle A_2O_2B_2$ as $2x$. Then, since $\triangle A_2O_2B_2$ is isosceles, $\angle O_2A_2B_2=90-x$. If we connect points $A_1$, $A_2$, $O_1$, and $O_2$ we get a quadrilateral. Since quadrilaterals have interior angles that add up to $360$ degrees, we know that $\angle A_1O_1O_2=180-2x$ (since the circles are tangent to line $s$). Then, following the same logic as above, $\angle O_1A_1B_1=x$. Now since we know that $\angle O_1A_1A_2$ and $\angle O_2A_2A_1$ are right, we get that $\angle A_2A_1P=90-x$ and $\angle A_1A_2P=x$. Therefore $\angle A_1PA_2=90$ and the statement is proven.