Triangle ABC is an isosceles triangle.
Solution:
Given data:
AB is parallel to CD and AC is the transversal.
∠ABC = 70° and ∠ACD = 55°
If two parallel lines are cut by a transversal, then alternate interior angles are congruent.
∠BAC ≅ ∠ACD
m∠BAC = m∠ACD
m∠BAC = 55°
Sum of the angles in a straight line = 180°
m∠ACD + m∠ACB + m∠ABC = 180°
55° + m∠ACB + 70° = 180°
m∠ACB + 125° = 180°
Subtract 125° from both sides, we get
m∠ACB = 55°
In triangle ABC,
∠BAC = 55° and ∠ACB = 55°
∠BAC = ∠ACB
Two angles in the triangle are equal.
Therefor triangle ABC is an isosceles triangle.
