Dear Mathlinkers, one of the two common external tangents is parallel to BC... Sincerely Jean-Louis

Dear Mathlikers, http://jl.ayme.pagesperso-orange.fr/vol49.html then Ortique 3, Problem 32... Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, one of the two common external tangents is parallel to BC... Sincerely Jean-Louis Yes, this is essentially what the problem is asking to prove. And in your second post you accidentally posted a link to a local file on your pc?

OK I have corrected Sincerely Jean-Louis

Let $I_B$, $I_C$, and $I$ be the incenters of $\triangle AHB$ and $\triangle AHC$. Observe that since $HI_B$ bisects $\angle AHB$, it follows that $HI_B\parallel CI$. Similarly, $HI_C\parallel CI$. Now, here is the crux. Construct parallelogram $HI_BJI_C$. Since $\angle ABH = \angle ACH$, we have $\angle AI_BH = \angle AI_CH$, so by Parallelogram Isogonality Lemma, we get that $AH$ and $AJ$ are isogonal w.r.t. $\angle I_BAI_C$. By angle chasing, this is equivalent to $AJ$ bisects $\angle BAC$. Thus, if we let $B' = I_BJ\cap AB$ and $C' = I_CJ\cap AC$, then quadrilaterals $AB'JC'$ and $ABIC$ are homothetic. In particular, this means $B'C'\parallel BC$. Moreover, $B'I_B$ bisects $\angle AB'C'$, so $B'C'$ is tangent to the incircle of $\triangle AHB$. Similarly, $B'C'$ is tangent to the incircle of $\triangle AHC$, so we are done.

Reflecting $ACH$ across $AH$, we reduce to the following problem. reduction wrote: Suppose $ABCD$ is a cyclic quadrilateral with $AB \perp CD$. Then prove the incircles of $ACD$ and $BCD$ share a tangent parallel to $AB$. To prove this, note by Fact 5 that the line $l$ formed by the incenters of $ACD$ and $BCD$ makes a $45^\circ$ angle with $AB$ and $CD$. Also note that the incircles share $CD$ as a tangent, so reflecting across $l$ gives the result.