trapezoid AD//BC, AB = BC = BD, altitude BK cuts AC at M, <CDM?

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trapezoid AD//BC, AB = BC = BD, altitude BK cuts AC at M, <CDM?

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Source: 2012 Oral Moscow Geometry Olympiad grades 8-9 p1

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#1 h Sep 8, 2019, 9:39 AM • 1 Y

Y by Adventure10

In trapezoid $ABCD$ , the sides $AD$ and $BC$ are parallel, and $AB = BC = BD$ . The height $BK$ intersects the diagonal $AC$ at $M$ . Find $\angle CDM$ .

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#2 h Sep 9, 2019, 12:57 PM • 2 Y

Y by Adventure10, Mango247

$\angle{CMD}=2\angle{CAD}=\angle{CBD}$
$\implies BMDC$ is a cyclic quadrilateral
$\implies \angle{CDM}=90^{\circ}$

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#3 h May 11, 2023, 5:06 PM

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The answer is $\angle CDM=90^{\circ}$ . To prove this, let's prove that the quadrilateral $BMDC$ is cyclic. Let $\angle CAD=\alpha$ . Then $\angle DBC=2\alpha$ , since $B$ is the circumcenter of triangle $ADC$ (from $AB=BC=BD$ ). The triangle $AMD$ is isosceles, so $\angle DMC=2\alpha$ . So we have $\angle MDC=2\alpha=\angle DBC$ , meaning the quadrilateral $BMDC$ is cyclic.

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