# If $H$ is the orthocenter of $\triangle ABC$, show that the circles on $AH$ and $BC$ cut orthogonally.

If $H$ is the orthocenter of $\triangle ABC$, show that the circles on $AH$ and $BC$ cut orthogonally.

If two circles cut orthogonally, it means that the angle between the tangents at the point of contact is 90 degrees.

I've noticed that the points of intersection of circles are the feet of altitudes from $B$ and $C$.

I think this looks like. But I'm not able to proceed. Please help.

Thanks.

• Does "the circle on $AH$" mean the circle whose diameter is $AH$? Commented Oct 4, 2016 at 7:12
• @mathlove Yes. $$Commented Oct 4, 2016 at 7:38 ## 2 Answers \qquad\qquad\qquad Let D,E,F be a point on BC,CA,AB such that AD\perp BC,BE\perp AC,CF\perp AB respectively. Then, we know that E,F exist on the two circles. Now consider the tangent lines at F for both circles. Then, let G be a point both on the tangent line for the circle on AH and on BC. Also, let I be a point both on the tangent line for the circle on BC and on AC. Now we have$$\angle{CFG}=\angle{FAH},\quad \angle{IFC}=\angle{FBC}$$Since$$\angle{FAH}+\angle{FBC}=\angle{BAD}+\angle{ABD}=90^\circ$$we have$$\angle{CFG}+\angle{IFC}=90^\circ.

• Why is $\angle CFG = \angle FAH$? Commented Oct 4, 2016 at 7:49
• @Dhruv: Because of the alternate segment theorem, which states that the angle between the tangent and chord equals the angle in the alternate segment. see here. I'll add a figure. Commented Oct 4, 2016 at 7:53

Let $AB\perp CI$, $D$ on $CI$, draw circles as shown. $AC$ meets the circles in $G$ and $H$, resp. Now $GB\parallel HD$ (Thales), hence $\triangle ABG$ is similar to $\triangle DCH$. Let the normals $FH$ and $EG$ meet in $J$. Consider $EIFJ$. Due to similarity the angles at $F$ and $E$ add up to $180^\circ$, so the angle at $J$ is $90^\circ$.

So regardless where you choose $D$, the normals and hence the tangents are perpendicular. Now if you choose $D$ to be the orthocenter, $G=H$ and the assertion follows.