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Source: Baltic Way 2016, Problem 18

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#1 h Nov 5, 2016, 10:48 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a parallelogram such that $\angle BAD = 60^{\circ}.$ Let $K$ and $L$ be the midpoints of $BC$ and $CD,$ respectively. Assuming that $ABKL$ is a cyclic quadrilateral, find $\angle ABD.$

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#4 h Nov 16, 2016, 9:46 PM • 3 Y

Y by StarLex1, Adventure10, Mango247

socrates wrote:

Let $ABCD$ be a parallelogram such that $\angle BAD = 60^{\circ}.$ Let $K$ and $L$ be the midpoints of $BC$ and $CD,$ respectively. Assuming that $ABKL$ is a cyclic quadrilateral, find $\angle ABD.$

Let $AB, LK$ intersect at $E$ , $EB=a,AB=2a$ and $EK=KL=b, BD=2b$ . I draw $BH$ perpendicular to $AB$ .

It is: $EB \cdot EA = EK \cdot EL \Leftrightarrow 3{a^2} = 2{b^2} \Leftrightarrow b = \frac{{a\sqrt 6 }}{2}$

We also have, $BH = a\sqrt 3 \Leftrightarrow BD = 2b = a\sqrt 3 \sqrt 2 = BH\sqrt 2$ , thus $\omega=45^0$ and finally $\boxed{\angle ABD=75^0}$

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This post has been edited 1 time. Last edited by george_54, Nov 16, 2016, 9:47 PM
Reason: typo

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#5 h Nov 17, 2016, 4:45 AM • 1 Y

Y by Adventure10

Yes that's exactly my solution!
Easy problem

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#6 h Apr 6, 2025, 2:43 PM

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Let $F$ be the intersection of $AL \cap BD$ . Since $\triangle BAF \sim \triangle FLD$ , we have $BF=2*FD$ . By some angle chase $\angle BAF=\angle FDA$ . So by the secant tangent theorem: $BF*BD=BA^2$ $\implies$ $\frac{2*BD^2}{3}=BA^2$ , and rest is just some bash.

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