Quad geo with circle (M,MA)

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Quad geo with circle (M,MA)

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Source: Iran MO 2024 3rd Round Geometry Exam P1

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#1 h Aug 27, 2024, 2:33 PM

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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 $\angle BZD=\angle BCD$ .

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Ianis

#2 h Aug 27, 2024, 3:29 PM

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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.

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#3 h Jan 11, 2025, 12:42 PM

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Diagram
For conciseness we define $\beta = \angle ABC$ .
We start with a "lemma".
Lemma
Proof
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 $\angle BZD = \angle BCD$ .

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cursed_tangent1434

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cursed_tangent1434

#4 h Feb 12, 2025, 10:49 AM

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Rather straightforward problem. Symmetry carries as all the way through. The following is the key claim.

Claim : We have the pairs of equal segments, $DX=DC$ and $BY=BC$ .

Proof : The first one is immediate. Simply note that $\triangle MXD \cong \triangle MCD$ since $MX=MC$ and
$\measuredangle XMD = \measuredangle XMC + \measuredangle CMD = 2\measuredangle XYC + \measuredangle CBD = 2\measuredangle DAC + \measuredangle CBD = \measuredangle DBC = \measuredangle DMC$ For the other, note that since $\overline{MB}$ is the external $\angle ABC-$ bisector, the reflection of $\overline{BC}$ across $\overline{MB}$ must be $\overline{AB}$ . Further, the reflection of point $C$ across $\overline{MB}$ must lie on $\overline{AB}$ but also on the circle centered at $M$ with radius $MC$ (or equivalently $MA$ ). Thus, the reflection of $C$ over $\overline{MB}$ is either $A$ or $Y$ of which the former can be immediately eliminated as it is on the same side of $\overline{MB}$ as point $C$ . Thus, $Y$ is the reflection of $C$ across $\overline{MB}$ which proves the claim.

Now note that $\triangle BZD \cong \triangle BCD$ since $BZ=BY=BC$ and,
$\measuredangle ZBD = \measuredangle ZBA + \measuredangle ABD = 2\measuredangle ZYB + \measuredangle ACD = 2\measuredangle XYA + \measuredangle ACD = 2\measuredangle XCA + \measuredangle ACD = \measuredangle XCD + \measuredangle XCA$ Further,
$\measuredangle DBC = \measuredangle DAC = \measuredangle XYC = \measuredangle XYA + \measuredangle AYC = \measuredangle XCA + \measuredangle AXC = \measuredangle XCA + \measuredangle XCD$ which implies that $\measuredangle ZBD = \measuredangle DBC$ .

Because of the congruent pair of triangles we can then conclude,
$\measuredangle BZD = \measuredangle DCB$ as desired.

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