Let ABCD be a parallelogram with AB>BC and ∠DAB less than ∠ABC. The perpendicular bisectors of sides AB and BC intersect at the point M lying on the extension of AD. If ∠MCD is equal to 15 degree, find the measure of ∠ABC

I don't usually do the Geometry problems, but since this one is d1, I'll give it a try.
In my sketch, A is to the lower right and the horizontal members of the parallelogram are the longer ones.
Since M is at the junction of two perpendicular bisectors, AM, BM and CM are all equal in length and if we also draw BM, two triangles are isoscelese:  triangle AMB and triangle BMC.
Call the obtuse angle at ABC angle x. So angle ADC is also x, and angle MDC is it's complement pi - x.
Since angle DCM is 15 degrees, angle AMC is pi - (pi - x + 15) = x - 15.  Or x = angle AMC + 15.

Angle AMC = angle AMB + angle BMC
AMB = pi - 2(pi-x) = 2x - pi
BMC = pi - 2(pi-x+15) = 2x - 30 - pi
AMB + BMC = AMC = x-15 = 2x - pi + 2x - 30 - pi
2*pi + 15 = 3x
x = (2/3)pi + 5

x = angle ABC = 125 degrees

Posted by Larry on 2020-12-23 09:08:48