2
$\begingroup$

Can you provide a proof for the following claim:

Claim.Given any $\triangle ABC$. The tangent lines to the circumscribed circle of $\triangle ABC$ are constructed at vertices $A$,$B$,$C$ . Let point $D$ be the intersection point of tangent lines that passes through vertices $A$ and $C$ , point $E$ the intersection point of tangent lines that passes through vertices $A$ and $B$ , point $F$ the intersection point of tangent lines that passes through vertices $B$ and $C$ , and let $H_1$,$H_2$,$H_3$ be the orthocenters of $\triangle ACD$, $\triangle AEB$ and $\triangle BFC$ respectively. Then line segments $AH_3$,$BH_1$ and $CH_2$ concur at the nine-point center of $\triangle ABC$.

enter image description here

GeoGebra applet that demonstrates this claim can be found here.

I don't know how to start the proof. All I know is that nine-point center of the triangle lies in the middle of the line segment whose endpoints are orthocenter and circumcenter , but I don't know how to use that fact. Any hints are welcomed.

$\endgroup$
1
$\begingroup$

In fact we can say more : $N_9$ is the midpoint of segments $AH_i$.

enter image description here

First we show that $BOCH_2$ is a rhombus. $\triangle BEC$ is isosceles so $H_2$ lies on its symmetry axis. So $BH_2C$ is isosceles as is $BOC$. Hence $OH_2$ is perpendicular to $BC$ at its midpoint $M$. Also $\angle OBC = 90 -A = 90 - \angle EBC = \angle H_2CB$.

Next we show $AHH_2O$ is a parallelogram. $AH \perp BC \Rightarrow AH || OH_2$. It is known that $2OM = AH$. So $AH = OH_2$.

We conclude $AH_2$ is bisected at intersection of diagonals of parallelogram $AHH_2O$. But midpoint of $OH$ is $N_9$. We're done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.