5
$\begingroup$

Let $\triangle ABC$ be a triangle. Let $Γ$ be its circumcircle, and let $I$ be it’s incenter. Let the internal angle bisectors of $∠A,∠B,∠C$ meet $Γ$ in $A',B',C'$ respectively. Let $B'C'$ intersect $AA'$ at $P$, and $AC$ in $Q$. Let $BB'$ intersect $AC$ in $R$. Suppose the quadrilateral $PIRQ$ is a kite; that is, $IP = IR$ and $QP = QR$. Prove that $\triangle ABC$ is an equilateral triangle.


I can prove upto isosceles triangle. How to prove equilateral?

$\endgroup$
  • $\begingroup$ If you can show the base angles of the isosceles triangle are $60$ degrees, you're done. $\endgroup$ – mattos Dec 13 '15 at 13:50
3
$\begingroup$

enter image description here

Hope you can complete your task with the aid of the above diagram.

Note: (1) $O$ will lie on the line $BIRB'$ making $BB'$ the diameter of the red circle; and (2) $QB' = QA$ provides further assistance in making each green marked angle (irrespective of their degree of shades) $= 30^0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.