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

circlesgeometry

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.

Diagram

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.

Best Answer

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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 $$\angle{CFG}+\angle{IFC}=90^\circ.$$

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