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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.

JMP
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Dheeraj Kumar
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  • Hi and welcome to the site! Since this is a site that encourages learning, you will get much more help if you show us what you have already done. Could you edit your question with your thoughts and ideas? – 5xum Dec 09 '14 at 15:39
  • Incentar is not collinear. Circum and orthocentar are collinear with the point that get where all three midlines intersect (I dont know how it is called in english). You are probably thinking about Eulers line? – CryoDrakon Dec 09 '14 at 22:04
  • And is it $\tan A \tan B$ or $\tan A \tan C$ ? – CryoDrakon Dec 09 '14 at 22:04

1 Answers1

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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 $\tan A\tan C=\frac{BE}{HE}=3$.

CryoDrakon
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