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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.

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  • $\begingroup$ Does "the circle on $AH$" mean the circle whose diameter is $AH$? $\endgroup$
    – mathlove
    Commented Oct 4, 2016 at 7:12
  • $\begingroup$ @mathlove Yes. $$ $\endgroup$ Commented Oct 4, 2016 at 7:38

2 Answers 2

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$\qquad\qquad\qquad$enter image description here

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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  • $\begingroup$ Why is $\angle CFG = \angle FAH$? $\endgroup$ Commented Oct 4, 2016 at 7:49
  • $\begingroup$ @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. $\endgroup$
    – mathlove
    Commented Oct 4, 2016 at 7:53
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enter image description here

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.

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