Prove the following:

Question

In the figure, line CD̅̅̅̅ is the median to line BC̅̅̅̅ and ∠A ≅ ∠B. Prove that line CD̅̅̅̅ is an altitude, and a perpendicular bisector to line AB̅̅̅̅ and it is also an angle bisector of ∠C.

Diagram: Triangle ACB (Upside Down) and line DC intersects them. A ≅ B, D ≅ C

Answer 1

∠A≅ ∠B

Therefore,

ΔACB is the isosceles triangle.

By definition, the median segment in a triangle is the segment connecting the vertex of the triangle and the midpoint of the base.

We know that the perpendicular to the base, drawn through the vertex of an isosceles triangle, will be the bisector to the base.

Since median and the perpendicular to the base connect the the same two points, it is the same segment; and median must be perpendicular to the base.

Therefore,

CD⊥AB

m∠ADC = m∠BDC = 90º

∠A and ∠ACD are complementary

m∠ACD = 90º - m∠A

∠B and ∠BCD are complementary

m∠BCD = 90º - m∠B

Therefore,

∠ACD ≅ ∠BCD → angles are congruent

It means that median CD will be the bisector of ∠ACB.