DM and CK are bisectors of angles D and C of trapezoid ABCD with bases AD and BC.

DM and CK are bisectors of angles D and C of trapezoid ABCD with bases AD and BC. Find the angle between these bisectors.

1. Let’s make a drawing.

2. Recall the properties of the trapezoid angles.

We are interested in the property about the corners adjacent to the lateral side of the trapezoid. The sum of these angles is 180 °. Means:

∠BCD + ∠ADC = 180 °.

3. Find the sum of the angles ECD and EDC.

Since the bisector CK divides ∠BCD in half, and the bisector DM divides ADC in half. Then:

∠ECD = ∠BCD: 2;

∠EDC = ∠ADC: 2.

Therefore, their sum will be equal to:

∠ECD + ∠EDC = ∠BCD: 2 + ∠ADC: 2;

∠ECD + ∠EDC = (∠BCD + ∠ADC): 2.

The sum ∠BCD + ∠ADC = 180 °, which means

∠ECD + ∠EDC = 180 °: 2;

∠ECD + ∠EDC = 90 °.

4. Determine the angle between the bisectors.

Consider a triangle EDC.

The sum of all three angles of any triangle is 180 °:

∠ECD + ∠EDC + ∠CED = 180 °.

Hence, ∠CED = 180 ° – (∠ECD + ∠EDC);

∠CED = 180 ° – 90 °;

∠CED = 90 °.

Answer: the angle between the bisectors DM and CK is 90 °.