4
$\begingroup$

Let $ABC$ be a triangle. Let $I$ be its inscribed circles' center. Let $D$ be the intersection point of lines $AI$ and $BC$. The perpendicular bisector of $AD$ intersects with $BI$ in $E$.

Thesis: $$ ED\perp CI $$

How to prove that using synthetic euclidean geometry?

$\endgroup$
1
$\begingroup$

The key is to show that $A,B,D,E$ are cyclic. To show this let the angle bisector of $\angle B$ cut the circumcircle of $\triangle ABD$ at $E^{'}$. Then, since $\angle ABE^{'}=\angle DBE^{'}$, we have $AE^{'}=DE^{'}$. Therefore the perpendicular bisector of $AD$ passes through $E^{'}$. Hence $E^{'}$ is the same as $E$, that is $A,B,D,E$ are cyclic as we desired. The remaining part is trivial.

Let $P$ be the intersection of $DE$ and $CI$ and $F$ the midpoint of $AD$. Using the cyclicity we've just proven, $$\angle AED=\pi-\angle ABD=2(\angle DAC+\angle ICA)$$ Because$ \triangle AED$ is isosceles, it's obvious that $\square PEFI$ is cyclic and the conclusion follows.

$\endgroup$

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.