In rhombus ABCD, the bisector of DCA is perpendicular to side AD. Find the larger corner of the diamond.

Let the value of the larger angle of the rhombus be X0. Angle DСВ = X0.

The diagonals of the rhombus divide the angle at the apex in half, then the angle is DCA = X / 2.

Since, by condition, CM is the bisector of the angle DCA, then the angle DCM = DCA / 2 = (X / 2) / 2 = X / 4.

The sum of the adjacent angles of the rhombus is 180, then the angle CDA = 180 – DCB = 180 – X.

Consider a triangle DСМ, the sum of the interior angles of which will be equal to:

(180 – X) + 90 + X / 4 = 180.

3 * X / 4 = 90.

X = 90 * 4/3 = 120.

Angle ВСD = 120.

Answer: The larger angle of the rhombus is 120.



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