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In $\triangle ABC$, line joining the circumcenter(O) and orthocenter(H) is parallel to side $AC$, then show that the value of $\tan A\tan C$ is 3.

Let $\perp$ from circumcenter cuts $AC$ at D and that from orthocenter cuts it at E.

Since line joining the circumcenter and orthocenter is parallel to side $AC$. $\perp$ distance from the circumcenter and orthocenter to $AC$ must also be equal.

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1 Answer

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Let point E be as you defined. We have that $\tan C=\frac{AE}{HE}$ and $\tan A=\frac{BE}{AE}$. From there we get that $\tan A\tan C=\frac{BE}{HE}$. Let BH intersect the outercircle of ABC in point E'. We have that EH=EE' (look it up on the internet why that is if you dont know it) and by using the intersecting secants theorem at point H we get: $R^2-OH^2=BH\cdot HE'=2BH\cdot HE$. We know that the angle BHO is $\frac{\pi}2$ so we have that $BO^2=HO^2+BH^2=R^2$. Putting that in the intersection secants theorem we get that $BH^2=2BH\cdot HE$ which leads to $BH=2HE \Rightarrow BE=3HE$. So now we have that $\tan A\tan C=\frac{BE}{HE}=3$.

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