As the figure shows, line $AB$ is perpendicular to line $CD$, and they intersect at $E$. Construct four circles with centers $A$, $B$, $C$, $D$ and radii $AE$, $BE$, $CE$, $DE$ respectively. Let $F,G,H,J$ be the intersections of those circles different than $E$. Prove that $F,G,H,J$ are on the same circle
I've tried to use the trigonometric functions to do it, which is a 'hard way'. But, I am sure that there is a much easier way to solve it. Can someone help me?
Image is for reference
Best Answer
$\angle FJE=\frac{\angle FAE}{2}=\frac{180°-2\angle FEA}{2}=90°-\angle FEA=\angle FED$
In the same way: $\angle HJE=\angle CEH$, $\angle HGE=90°-\angle CEH$, $\angle FGE=90°-\angle FED$.
Then $\angle FJH + \angle FGH = \angle FJE+\angle HJE+\angle HGE+\angle FGE=$ $\angle FED+\angle CEH+90°-\angle CEH+90°-\angle FED=180°$.
Then $FJHG$ is cyclic quadrilateral.