$ABC$ is a right-angled triangle, with $\angle ABC = 90^{\circ}$. $B'$ is the reflection of $B$ over $AC$. $M$ is the midpoint of $AC$. We choose $D$ on $\overrightarrow{BM}$, such that $BD = AC$. Prove that $B'C$ is the angle bisector of $\angle MB'D$. NOTE: An important condition not mentioned in the original problem is $AB<BC$. Otherwise, $\angle MB'D$ is not defined or $B'C$ is the external bisector.
Problem
Source: OMCC 2017
Tags: OMCC, OMCC 2017, CENTRO, geometry, geometric transformation, reflection, angle bisector
Mewi
23.06.2017 23:18
$BD=AC$ and M is the midpoint of $AC$, so $MB=MC=MB'=MD$. Then $\angle B'MB=180-2\angle MB'B$, and $2\angle MB'D=\angle B'MB$. So $180-2\angle MB'B=2\angle MB'D$ $\rightarrow$ $\angle MB'B+\angle MB'D=90$ Due to the two right angles, we have that $B'D\parallel MC$, finally $\angle MB'C=\angle DB'C$ That is the same as saying that $B'C$ is the angle bisector of $\angle MB'D$.
Wave-Particle
24.06.2017 01:25
Note that $M$ is the circumcenter of the circle that contains points $B,A,B',D,C$.
Thus, it suffices to show that $\angle MB'C = \angle MBC$. However, this is trivial since $\angle B'BC = \angle BB'C$ and $\angle MBB' = \angle MB'B$ by the fact that $MB=MB'$. $\blacksquare$
csaltachin
29.05.2018 06:26
It's easy to see that $M$ is the center of the circle that contains $A,B,C,B'$ and $D$, $AB=AB'$ because of the reflection, and $AB=CD$ since $\bigtriangleup ABM\cong\bigtriangleup DCM$. Thus $\Box AB'DC$ is an isosceles trapeze, and $\angle CB'D=\angle ADB'=\frac{\angle BDB'}{2}=\frac{\angle MDB'}{2}$ since $A$ is the midpoint of $\overarc{BAB'}$ and $B-M-D$.
Yet notice that $\bigtriangleup B'MD$ is isosceles on $M$, meaning $\angle MDB'=\angle MB'D$ and $\angle CB'D=\frac{\angle MB'D}{2}$ as desired.
AlastorMoody
21.02.2019 10:38
Easy to see, Points $A,B,C,D,B'$ lie on a circle, and clearly, $BB'$ and $B'M$ are isogonal WRT $\angle AB'C$, hence, $\angle AB'B=$ $\angle MB'C=$ $\angle ABB'=\angle MBC=\angle DB'C$
ilikemath40
16.09.2021 05:30
[asy][asy] import olympiad; size(7cm); pair A, B, B1, M, D, C, E, F; M = (4, 0); draw(circle(M, 4)); A = (0, 0); C = (8, 0); F = (1.47, 0); B = (1.47, 3.1); B1 = (1.47, -3.1); draw(A--B--C--A); draw(A--B1--C--A); D = (6.53, -3.1); E = (6.53, 3.1); draw(B--D); draw(B1--E); draw(B--B1); draw(B1--D); draw(A--B1); dot(A^^B^^B1^^M^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, N); label("$C$", C, NE); label("$D$", D, SE); label("$B'$", B1, SW); label("$E$", E, N); label("$M$", M, N); label("$F$", F, NE); [/asy][/asy] $\textbf{Claim:}$ $B'$ lies on circle $M$. Proof. Notice that if we reflect $\triangle{ABC}$ over the segment $AC$ we get $\triangle{AB'C}$. Notice that the angles of the triangle are preserved. Since $\angle{AB'C}=90^\circ$ we know that $B'$ must lie on the circle. $\square$ $\textbf{Claim:}$ $D$ lies on circle $M$. Proof. Trivial because diameter. $\square$ Notice that $BF=B'F$ and that $\angle{BFC}=\angle{B'FC}=90^\circ$. It follows that $\triangle{BFC} \cong \triangle{B'FC}$. So we have \[ \angle{BCF}=\angle{FCB'} \implies \widehat{AB}=\widehat{AB'} \implies \widehat{CD}=\widehat{EC} \]Thus we have $\angle{EB'C}=\angle{CB'D}$ because of inscribed angles.
OlympusHero
16.09.2021 08:03
Denote $\angle A = a$. Then angle chase to get $\angle B'DM = 180-2a$ and $\angle DMC = 180-2a$, so $B'D$ is parallel to $MC$. This means $\angle DB'C=\angle MCB'$. It now remains to show $\angle MCB'=\angle MB'C$, or $MB'=MC$. But $MB=MC=MB'$, so this is true.
ilikemath40
16.09.2021 15:14
OlympusHero wrote:
Denote $\angle A = a$. Then angle chase to get $\angle BDM = 180-2a$ and $\angle DMC = 180-2a$, so $BD$ is parallel to $MC$. This means $\angle DB'C=\angle MCB'$. It now remains to show $\angle MCB'=\angle MB'C$, or $MB'=MC$. But $MB=MC=MB'$, so this is true.
What is $\angle{BDM}$? Its a straight angle. Also $BD \parallel MC$ is not true.
OlympusHero
16.09.2021 17:40
Sorry, I was using $B$ sometimes instead of $B'$
MathLuis
16.09.2021 18:08
Let $B'M$ meet $(ABC)$ at $F$. Note that $\angle AB'C=90$ thus $B'$ lies on $(M)$ and since $M$ is center $D$ is also on $(M)$ becuase $BD$ has to be a diameter. $AC$ and $BD$ meet on a common midpoint thus $ABCD$ is a paralelogram but since its also cyclic we have that $ABCD$ is a rectangle. $AC$ and $B'F$ meet on a common midpoint thus $AB'CF$ is a paralelogram but since its also cyclic we havd $AB'CF$ rectangle. Note that $AB=AB'$ thus we have $DC=AB=AB'=CF$ and we are done