anglechasing starting with a parallelogram (Kharkiv City VIII 2019 Ukr)

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anglechasing starting with a parallelogram (Kharkiv City VIII 2019 Ukr)

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#1 h Mar 22, 2020, 12:37 PM

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Let $ABCD$ be parallelogram with $\angle ADC = 40^o$ . Let K be a point such that the segment $AK$ intersects the side $BC$ , $AK = BC$ and $\angle BAK = 80^o$ . Let $L$ be a point such that the segment $CL$ intersects the side $AD$ , $CL = AB$ and $\angle BCL = 80^o$ . Find the angles of the triangle $BKL$ . Justify your answer.

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#2 h Mar 24, 2020, 12:56 PM

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Notice $AKD, CDL$ are equilateral triangles, take $M$ reflection of $K$ across the midpoint of the diagonals of parallelogram, notice that triangles $BCM, ALM$ are equilateral and easily $\triangle BML\cong\triangle BCK$ giving $BL=BK$ and $m(\widehat{KBL})=60^\circ$ .
Remark: However it's possible to have been requested a vectorial solution.

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#3 h Aug 3, 2020, 12:57 PM • 1 Y

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