angle chasing , danube junior geometry 2014

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Source: Danube 2014 Junior P3

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#1 h Sep 7, 2018, 8:38 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle with $\angle A<90^o, AB \ne AC$ . Denote $H$ the orthocenter of triangle $ABC$ , $N$ the midpoint of segment $[AH]$ , $M$ the midpoint of segment $[BC]$ and $D$ the intersection point of the angle bisector of $\angle BAC$ with the segment $[MN]$ . Prove that $<ADH=90^o$

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#2 h Sep 7, 2018, 10:40 AM • 2 Y

Y by Adventure10, Mango247

Let $O$ be the circumcenter of $ABC$ .
Since $N$ is the midpoint of $AH$ , we want to prove that $D$ is on the Thales circle around $N$ which is equivalent to $\angle DAN=\angle NDA$ .
But since $\angle BAO=\angle HAC=90^\circ-\gamma$ , $AD$ is also the angular bisector of $OAH$ . Hence it suffices to prove that $\angle NDA=\angle OAD$ i.e. that $OA$ and $MN$ are parallel. But since clearly $AN \parallel OM$ (both are perpendicular to $BC$ ) this is equivalent to $OAMN$ being a parallelogram i.e. it suffices to check that $AN$ and $OM$ have the same length i.e. that $OM$ is half as long as $AH$ .
Let $X$ be the point of reflection of $H$ by $M$ . It is well-known that $X$ lies on the circumcircle. Moreover, we have $HC \parallel BX$ so $\angle XBA=90^\circ$ so $X$ is opposite to $A$ on the circumcircle. But then $O,X,A$ are collinear so triangles $AHX$ and $OMX$ are similar whence finally $\frac{AH}{OM}=\frac{AX}{OX}=2$ .

This post has been edited 2 times. Last edited by Tintarn, Sep 7, 2018, 11:13 AM

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#3 h Sep 8, 2018, 4:04 PM • 4 Y

Y by amar_04, Adventure10, Mango247, MS_asdfgzxcvb

Let $BE,CF$ altitudes of $\triangle ABC$ . Clearly $EF$ is radical axis of the circles of diameters $(AH), (BC)$ , thus $EF\bot MN$ . Since $MN$ is perpendicular bisector of $EF$ , we get its intersection with the bisector of $\angle BAC$ lies onto the circle $\odot (AEF)$ .

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#4 h Mar 4, 2025, 3:15 PM

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Another solution
We know that by 9 point circle ,foot of perpendiculars ,midpoints of sides and midpoints of sides are concyclic. Let foot of perpendicular from A be P and from C be Q
Now P,M,H,N are concyclic. rest is just angle chasing

This post has been edited 1 time. Last edited by Gggvds1, Mar 4, 2025, 3:16 PM

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#5 h Mar 6, 2025, 5:51 AM

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All is angle chasing:
Denote $H’$ the reflection of $H$ in $M$ , $D’$ the orthogonal projection of $H$ into the bisector of $\angle ABC$ . We aim to show that $I, D’, M$ are collinear which should do it.
By a common result, $AH’$ passes through the circumcenter of $\triangle ABC$ thus
$\angle AID’ = 2\angle AHD’ = 180^{\circ} - 2\angle HAD’ = 180^{\circ} - \angle HAH’ = \angle AIM$ .
Thus, $I, D’, M’$ are collinear and we are done.

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