OJBM

Geometry Difficulty 6.1 National Olympiad Prove it JBMO

Problem:
Let $ABC$ be an isosceles triangle such that $AB = AC$ and $\angle \frac{A}{2} < \angle B$ . On the extension of the altitude $AM$ we get the points $D$ and $Z$ such that $\angle CBD = \angle A$ and $\angle ZBA = 90^{\circ}$ . $E$ is the foot of the perpendicular from $M$ to the altitude $BF$ and $K$ is the foot of the perpendicular from $Z$ to $AE$ . Prove that $\angle KDZ = \angle KBD = \angle KZB$ .

Solution

Solution:
The points $A, B, K, Z$ and $C$ are co-cyclic.
Because $ME \parallel AC$ so we have
$\angle KEM = \angle EAC = \angle MBK$
Therefore the points $B, K, M$ and $E$ are co-cyclic. Now, we have
$\begin{aligned} & \angle ABF = \angle ABC - \angle FBC \\ & = \angle AKC - \angle EKM = \angle MKC \end{aligned}$
Also, we have
$\begin{aligned} & \angle ABF = 90^{\circ} - \angle BAF = 90^{\circ} - \angle MBD \\ & = \angle BDM = \angle MDC \end{aligned}$
From (1) and (2) we get $\angle MKC = \angle MDC$ and so the points $M, K, D$ and $C$ are co-cyclic.
Consequently,
$\angle KDM = \angle KCM = \angle BAK = \angle BZK,$
and because the line $BD$ is tangent to the circumcircle of triangle $ABC$ , we have
$\angle KBD = \angle BAK$

Figure 5
Finally, we have
$\angle KDZ = \angle KBD = \angle KZB$

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