A rectangle $ABCD$ is given. Segment $DK$ is equal to $BD$ and lies on the half-line $DC$. $M$ is the midpoint of $BK$. Prove that $AM$ is the angle bisector of $\angle BAC$. Proposed by S. Berlov
Problem
Source: Tuymaada 2012, Problem 2, Day 1, Juniors
Tags: geometry, rectangle, symmetry, angle bisector, geometry proposed
applepi2000
22.07.2012 01:02
Let $\angle BDK=\angle BAC=x$. Then since $\angle CBK=(\frac{\pi}{2}-x)-(\frac{\pi}{2}-\frac{x}{2})=\frac{x}{2}$ and $M$ is the center of circle $BCK$, we have $\angle BAC=x=\angle CMK=\pi-\angle CMB$. So $ABMC$ is cyclic, and $\angle CAM=\angle CBM=\frac{x}{2}$ as desired.
professordad
22.07.2012 01:12
Let there be a point $L$ on $\overline{AB}$ such that $AC = AL$. By symmetry, $\triangle BLC \cong \triangle CKB$; also, since $BM = MK$, $LM = MC$. But $AC = AL$, so by Angle Bisector Theorem on $\triangle ACL$ we're done.
sunken rock
22.07.2012 14:46
Obviously, $MD$ is angle bisector of $\angle BDC$ and $MD\bot BK$, so $ABMD$ is cyclic, and $\angle BAM=\angle BDM=\frac{\angle BDC}{2}=\frac{\angle BAC}{2}$, done. Best regards, sunken rock
Marius Mainea
02.08.2012 18:52
Let E on the half line CE. Then AB=KE ,so CE=DK=DB=AC. Result $\angle{BAE}=\angle{AEC}=\angle{CAE}$
Mahdi_Mashayekhi
02.01.2022 16:02
∠MCA = 180 - ∠KCM - ∠ACD = 180 - ∠ABD - ∠DBK = 180 - ∠ABM ---> ABMC is cyclic. ∠CAM = ∠CBM and ∠BAM = ∠BCM. we're Done.