DA is the median of an isosceles triangle BDC with base CB: angle D = 120. Find the angles of triangle ADC.

1) Since the triangle ВСD is isosceles and ∠ Д = 120 °, then we will find 2 second angles.

It is known that CB is a base, then: ∠ С = ∠ V.

The sum of two equal angles B and C is equal to:

∠ B + ∠ C = 180 ° – ∠ D;

∠ В + ∠ С = 180 ° – 120 °;

∠ В + ∠ С = 60 °;

2 * ∠ С = 60 °;

∠ С = 60 ° / 2;

∠ С = 30 °.

2) The median AD divides the BC side in half, it is also the height and the bisector.

Hence, ∠ A = 90 °.

The bisector divides the angle in half, which means: ∠ CDA = 1/2 * ∠ D = 1/2 * 120 ° = 60 °.

3) Consider the triangle ACD and find its angles.

∠ С = 30 °;

∠ А = 90 °;

∠ D = 60 °.