Prove that the sides opposite to equal angles of a triangle are equal.


Answer:


Step by Step Explanation:
  1. Let ^@ ABC^@ be an isosceles triangle with ^@\angle B = \angle C.^@
      A B C
  2. We need to prove that ^@ AB = AC ^@
  3. Let us construct an angle bisector ^@AD^@ of angle ^@A.^@
      A B CD
    In ^@\triangle ABD^@ and ^@\triangle ACD,^@ we have @^ \begin{aligned} & \angle B = \angle C && \text{[Given]} \\ & \angle BAD = \angle CAD && \text{[By construction]} \\ & \ AD = AD && \text{[Common]} \\ & \therefore \space \triangle ABD \cong \triangle ACD && [\text{By AAS citerion}] \end{aligned} @^
  4. As the corresponding parts of congruent triangles are equal, we have @^ AB = AC. @^
  5. Thus, the sides opposite to equal angles of a triangle are equal.

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