5
$\begingroup$

enter image description here

$D$ is the incenter of $\triangle ABC$. $DE \perp BC$ ($E \in BC$). $AE \cap \bigcirc(A, B, C) = F$ ($F \not\equiv A$). $G$ is the midpoint of the larger arc of $BC$. $GF \cap \bigcirc(B, C, D) = {H}$ ($GH < GF$). Prove that the excenter of $A$ in $\triangle ABC$, the midpoint of $BC$ and $H$ are collinear.

Let the midpoint of $BC$ and the excenter of $A$ in $\triangle ABC$ be respectively $I$ and $K$.

What I am trying to prove is that $HI \parallel AE$ and $HK \parallel AF$. (Perhaps $EFIH$ and $AFKH$ are parallelograms.) But I don't exactly know how.

$\endgroup$
1
+50
$\begingroup$

enter image description here

Extend $DE$ to meet the circumcircle of $BDC$ again at $I$. By Power of a Point, $DE\times IE=BE\times EC=AE\times EF$, so $A,D,I,F$ are concyclic. Therefore, $\angle DAF=\angle EIF$. enter image description here

Now, extend IF to meet the circumcircle of $ABC$ at $G'$. Construct a line through $G'$ parallel to $DE$ (i.e. perpendicular to $BC$. Let this line meet $BC$ at $M$ and circumcircle of $ABC$ at $N$. Then we have $\angle FG'N=\angle FID=\angle ADF$ Therefore $AD$ meets $G'N$ at $N$. Since $AD$ meets circumcircle $ABC$ at the midpoint of arc $BC$, this midpoint is $N$ and therefore $G'=G$ and $M$ is the midpoint of $BC$. In essence, we have shown that $G,F,I$ are collinear.

enter image description here

Now, it is time to clean up. Let the $A$-excenter be $P$. Then it is very well known that $A,D,P$ are collinear and $DP$ is the diameter of circle $BDC$. This means that $\angle PID =90^{\circ}$ so $PI$ is parallel to $BC$. Therefore arc $BI$ equals to arc $CP$ in circle $BDC$. Therefore $\angle BHI=\angle CHP$. It suffices to show that $\angle BHI=\angle CHM$.

But it is equally well-known that $GB,GC$ are tangents to the circumcircle of $BDHC$. Since $G,H,I$ collinear, $GH$ is actually the $H$-symmedian of $\triangle BHC$. Therefore $\angle BHI=\angle CHM$ and we are done.

$\endgroup$

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.