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, 2009

Geometry Difficulty 5.0 AIME Prove it JBMO

Problem:
Let $ABCD$ be a parallelogram with $AC > BD$ , and let $O$ be the point of intersection of $AC$ and $BD$ . The circle with center at $O$ and radius $OA$ intersects the extensions of $AD$ and $AB$ at points $G$ and $L$ , respectively. Let $Z$ be the intersection point of lines $BD$ and $GL$ . Prove that $\angle ZCA = 90^{\circ}$ .

Solution

Solution:
From the point $L$ we draw a parallel line to $BD$ that intersects lines $AC$ and $AG$ at points $N$ and $R$ respectively. Since $DO = OB$ , we have that $NR = NL$ , and point $N$ is the midpoint of segment $LR$ .

Let $K$ be the midpoint of $GL$ . Now, $NK \parallel RG$ , and
$\angle AGL = \angle NKL = \angle ACL$
Therefore, from the cyclic quadrilateral $NKCL$ we deduce:
$\angle KCN = \angle KLN$
Now, since $LR \parallel DZ$ , we have
$\angle KLN = \angle KZO$

It implies that quadrilateral $OKCZ$ is cyclic, and
$\angle OKZ = \angle OCZ$
Since $OK \perp GL$ , we derive that $\angle ZCA = 90^{\circ}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.