In a parallelogram ABCD BC = 2AB. M is the middle of AD. A perpendicular is drawn from M, and a perpendicular is drawn from the vertex C to the continuation AB (point E). Prove that angle DME = 3 angles AEM.
Let’s connect the points M and C and consider the triangle MCD.
Because | DM | = | DC | by the condition of the problem, the triangle MCD is isosceles and the angles CMD and MCD are equal. We denote the angle CMD = MCD = a. Then the angle is МDC = 180 – 2a.
Then, since AB is parallel to CD, then the angle BAD = 180 – MDC = 180 – (180 -2a) = 2a.
Triangle ABM is also isosceles because | AB | = | AM |.
Therefore, the angles ABM and AMB are equal. Then the angle AMB = 0.5 (180 – BAD) = 0.5 (180-2a) = 90 – a
Then the angle BMC = 180 – AMB – CMD = 180 – (90 -a) – a = 90, i.e. BMC straight.
So, triangles BMC and BEC are rectangular with a common hypotenuse, and therefore points B M E C lie on a circle with a center at point O and radius | AM |.
Note that the angle EBC and EMC are based on the same arc, which means they are equal. But EBC = BAD = 2a.
So EMC = 2a. And then DME = EMC + CMD = 2a + a = 3a
Note that the angle BEM and BCM rest on the same arc, which means they are equal. BEM = BCM = CMD = a
So, DME = 3a = 3AEM, as required.