Prove that ∠A≅∠C in △ABC below.

Answer 1

To prove that ∠A≅∠C in △ABC, we need to show that they are congruent (or equal) angles.

Given that △ABC is a triangle, we know that the sum of the angles in any triangle is 180°.

Let's assume that m∠A < m∠C. This means that ∠C is the larger angle.

Since ∠A and ∠C are angles of the triangle △ABC, the other angle ∠B is equal to 180° - (m∠A + m∠C).

If we switch the assumption and assume that m∠A > m∠C, then ∠A becomes the larger angle and ∠B becomes equal to 180° - (m∠C + m∠A).

In either case, we can see that ∠B, the remaining angle in the triangle, is equal to both 180° - (m∠A + m∠C) and 180° - (m∠C + m∠A).

This implies that ∠B is equal to itself.

However, according to the Transitive Property of Equality, if ∠B is equal to itself and ∠B is equal to 180° - (m∠A + m∠C) and ∠B is equal to 180° - (m∠C + m∠A), then 180° - (m∠A + m∠C) must also be equal to 180° - (m∠C + m∠A).

Simplifying this equation:

180° - (m∠A + m∠C) = 180° - (m∠C + m∠A)
180° - m∠A - m∠C = 180° - m∠C - m∠A

By subtracting 180° from both sides:

- m∠A - m∠C = - m∠C - m∠A

By adding m∠C to both sides:

- m∠A = - m∠A

Since the negative of a number is equal to itself, we have:

m∠A = m∠A

This shows that ∠A is congruent to itself.

According to the Reflexive Property of Congruence, any angle is congruent to itself.

Therefore, we can conclude that ∠A ≅ ∠C.