point lies on median, isosceles, equal segments, angle bisector related

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Source: 2016 Kurchatov Olympiad 9 p3 - Russia

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#1 h Mar 8, 2021, 3:25 PM

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You are given an isosceles triangle $ABC$ . On the lateral side $AB$ , a point $M$ is marked such that $CM = AC$ . Then, on the lateral side $BC$ , a point $N$ was marked such that $BN = MN$ , and the angle bisector $NH$ was drawn at triangle $CNM$ . Prove that $H$ lies on the median $BK$ of the triangle $ABC$ .

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#2 h Mar 8, 2021, 11:42 PM

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Funny.

Note that the problem conditions translates to $ACNM$ cyclic and $NA = NC$ . Furthermore, let $CM \cap \ell$ where $\ell$ is the angle bisector of $\angle CNM$ intersect at $P$ . If we let $\angle B = \angle C = \theta$ , then $\angle PMN = \angle CAN = \theta \text{ and } \angle PNM = \tfrac12(180^{\circ} - \angle BNM) = \theta$ as well. Hence, $BP$ is the $B$ -symmedian of $\triangle BMN$ and since $MN, BC$ antiparallel we are done.

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#3 h Mar 9, 2021, 4:43 AM

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jj_ca888 wrote:

Funny.

Note that the problem conditions translates to $ACNM$ cyclic and $NA = NC$ . Furthermore, let $CM \cap \ell$ where $\ell$ is the angle bisector of $\angle CNM$ intersect at $P$ . If we let $\angle B = \angle C = \theta$ , then $\angle PMN = \angle CAN = \theta \text{ and } \angle PNM = \tfrac12(180^{\circ} - \angle BNM) = \theta$ as well. Hence, $BP$ is the $B$ -symmedian of $\triangle BMN$ and since $MN, BC$ antiparallel we are done.

I'm pretty sure $ACNM$ isnt cyclic...

Anyway, my solution

Let $\angle BAC = \angle BCA = x$ , then, $\angle AMC = \angle MAC = x$ , $\angle ABC = 180 - 2a$ and so $\angle BMN = \angle MBN = 180 - 2x$ which means $\angle CMN = x$ . Since $\angle ACM = 180 - 2x$ , $\angle MCN = 3x-180$ which means $\angle MNC = 360-4x$ . Since $NH$ is angle bisector, $\angle MNH = 180 - 2x$ which means $\angle NHM = x$ .

But since $\angle NHM = \angle HMN = x$ , we get that $NH = NM = NB$ and since $\angle BNH = 2x$ , we get $\angle HBC = 90 - x$ , which means $H$ lies on the median/angle bisector $BK$

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