Let $ABCD$ be a cyclic quadrilateral with circumcircle $\Gamma$. Let $M$ be the midpoint of the arc $ABC$. The circle with center $M$ and radius $MA$ meets $AD, AB$ at $X, Y$. The point $Z \in XY$ with $Z \neq Y$ satisfies $BY=BZ$. Show that $\angle BZD=\angle BCD$.
Problem
Source: Iran MO 2024 3rd Round Geometry Exam P1
Tags: geometry
Ianis
27.08.2024 18:29
We have that $DM$ is the angle bisector of $\angle ADC$ and that $BM$ is the exterior angle bisector of $\angle ABC$, so from $MX=MC=MY$ we get that $X$ and $C$ are symmetric wrt $MD$ and that $Y$ and $C$ are symmetric wrt $BM$. Let $Z'$ be the reflection of $C$ wrt $DB$. Then $X,Y,Z'$ lie on the Steiner Line of $C$ wrt $BMD$ and $\angle BZ'D=\angle BCD$. Hence $Z=Z'$, so $BY=BC=BZ$ (and also $DX=DC=DZ$). Done.
Ihatecombin
11.01.2025 15:42
[asy][asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -10.499716928401632, xmax = 20.780723164619697, ymin = -10.54620185809838, ymax = 23.41541881431965; /* image dimensions */
/* draw figures */
draw(circle((5,5), 5), linewidth(0.4));
draw((2.8097164274564284,9.49473668548512)--(0.7631615534283691,2.3449670477275637), linewidth(0.4));
draw((0.7631615534283691,2.3449670477275637)--(9.236838446571632,2.3449670477275637), linewidth(0.4));
draw((9.236838446571632,2.3449670477275637)--(2.8097164274564284,9.49473668548512), linewidth(0.4));
draw((0.7631615534283691,2.3449670477275637)--(3.667585328101434,0.18080181543553486), linewidth(0.4));
draw((3.667585328101434,0.18080181543553486)--(9.236838446571632,2.3449670477275637), linewidth(0.4));
draw(circle((5,0), 4.8424859811129695), linewidth(0.4) + red);
draw((0.2914734700956391,-1.1311270461192395)--(3.667585328101434,0.18080181543553486), linewidth(0.4));
draw((2.8097164274564284,9.49473668548512)--(3.667585328101434,0.18080181543553486), linewidth(0.4));
draw(circle((2.8097164274564284,9.49473668548512), 7.436907470542297), linewidth(0.4) + green);
draw((3.667585328101434,0.18080181543553486)--(6.127832845835589,2.83908524032286), linewidth(0.4));
draw((6.127832845835589,2.83908524032286)--(2.8097164274564284,9.49473668548512), linewidth(0.4));
draw((0.2914734700956391,-1.1311270461192395)--(7.781466078697126,3.9639774129443377), linewidth(0.4));
/* dots and labels */
dot((2.8097164274564284,9.49473668548512),linewidth(3pt) + dotstyle);
label("$B$", (2.9955586545875708,9.74139785937239), NE * labelscalefactor);
dot((0.7631615534283691,2.3449670477275637),linewidth(3pt) + dotstyle);
label("$C$", (0.9399868770461689,2.591582980968595), NE * labelscalefactor);
dot((9.236838446571632,2.3449670477275637),linewidth(3pt) + dotstyle);
label("$A$", (9.430392045151958,2.591582980968595), NE * labelscalefactor);
dot((3.667585328101434,0.18080181543553486),linewidth(3pt) + dotstyle);
label("$D$", (3.8445991713981496,0.4466385174474565), NE * labelscalefactor);
dot((5,0),linewidth(3pt) + dotstyle);
label("$M$", (5.185189461099064,0.2678931454873616), NE * labelscalefactor);
dot((0.2914734700956391,-1.1311270461192395),linewidth(3pt) + dotstyle);
label("$X$", (0.44843710415583377,-0.8492654292632315), NE * labelscalefactor);
dot((7.781466078697126,3.9639774129443377),linewidth(3pt) + dotstyle);
label("$Y$", (7.955742726480953,4.2449776715994725), NE * labelscalefactor);
dot((6.127832845835589,2.83908524032286),linewidth(3pt) + dotstyle);
label("$Z$", (6.302348035849826,3.1278190968488797), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy][/asy]
For conciseness we define \(\beta = \angle ABC\). We start with a "lemma".
\(BY = BC\)
This is fairly well known but I'll include it for completeness, we have
\[\angle BYC = 180 - \angle CYA = \dfrac{1}{2}\angle CMA = 90 - \dfrac{\beta}{2}\]However we have \(\angle YBC = \beta\) whence the conclusion immediately follows
We wish to prove that \(\triangle BZD \cong \triangle BCD\), this clearly finishes the problem and due to the previous lemma it suffices to show that \(\angle CBD = \angle DBZ\). We shall chase \(\angle BYZ\) \begin{align*} \angle BYZ &= \angle BYX \\ &= \angle BYC + \angle CYX \\ &= 90 - \dfrac{\beta}{2} + \angle CAD && \tag{\(CYAX\) is cyclic} \\ &= 90 - \dfrac{\beta}{2} + \angle CBD \\ \end{align*}We thus obtain that \[\angle ZBY = 180 - 2\angle BYZ = \beta - 2\angle CBD\]However it is clear that \[\angle ZBC = \beta - \angle ZBY = 2\angle CBD\]We now obtain \[\angle DBZ = \angle ZBC - \angle CBD = \angle CBD\]Thus we are done, we obtain that \(\triangle BZD \cong \triangle BCD\) and thus \(\angle BZD = \angle BCD\).