Question

Given: In triangle ABC, ∠B ≅ ∠C Prove: AB ≅ AC Complete the paragraph proof. We are given that ∠B ≅ ∠C. Assume segment AB is not congruent to . If AB > AC, then m∠C > m∠B by the . If AB < AC, then m∠C < m∠B by the converse of the triangle parts relationship theorem. But by the definition of congruent, we know the measure of angle B equals the measure of by the given statement. Therefore, we have a contradiction: AB = AC, and AB ≅ AC.

Answer 1

Answer:    

Given: Here ABC is a triangle,

In which $\angle B \cong \angle C$

Prove: $AB \cong AC$

Let us assume AB is not congruent to AC.

If $AB > AC$,

Then $m\angle C > m\angle B$ ( By the converse of the triangle parts relationship theorem )

If  If AB < AC, then $m\angle C < m\angle B$  ( by the converse of the triangle parts relationship theorem. )

But, by the definition of congruent,

We know $\angle B \cong \angle C$ ( by the given statement.)

Therefore, we have a contradiction,

And, $AB \cong AC$  and $AB = AC$.

Answer 2

Answer:

AB ≅ AC

Step-by-step explanation:

Given information: In triangle ABC, ∠B ≅ ∠C.

To prove: AB ≅ AC

Let us assume side AB is not congruent to side AC.

If AB > AC, then ∠B < ∠C ( By the converse of the triangle parts relationship theorem ).

If AB < AC, then ∠B > ∠C ( By the converse of the triangle parts relationship theorem ).

It is given that ∠B ≅ ∠C, so by the definition of congruent segment.

$\angle B=\angle C$

Therefore, we have a contradiction:

AB = AC

By the definition of congruent segment.

AB ≅ AC

Hence proved.