Let $ABCD$ be a parallelogram with $AB>BC$ and $\angle DAB$ less than $\angle ABC$. The perpendicular bisectors of sides $AB$ and $BC$ intersect at the point $M$ lying on the extension of $AD$. If $\angle MCD=15^{\circ}$, find the measure of $\angle ABC$
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Tags: geometry, parallelogram, perpendicular bisector
atmchallenge
29.04.2015 06:27
In $\LaTeX$: Let $ABCD$ be a parallelogram with $AB>BC$ and $\angle DAB$ less than $\angle ABC$. The perpendicular bisectors of sides $AB$ and $BC$ intersect at the point $M$ lying on the extension of $AD$. If $\angle MCD=15^{\circ}$, find the measure of $\angle ABC$.
Dwu123
29.04.2015 06:30
What does perpendicular bisector of sides AB and BC mean?
fprosk
29.04.2015 06:33
The perpendicular bisector of AB and the perpendicular bisector of BC. It's the line that is perpendicular to the segment that cuts the segment in half.
WorstIreliaNA
29.04.2015 06:34
The perpendicular bisector of AB is the line that is perpendicular to AB and bisects AB....
Omez
29.04.2015 06:48
Name x=<ABC. M is the circumcentre of ABC so <AMC=2(180-x) and we have also <MDC=180-x. So from <MCD+<MDC+AMC=180 we obtain 180-x+2(180-x)+15=180 so x=125. Done