We prove “If AB=AC, then ∠B=∠C” by using congruence.
Example ① (Proof of the Base Angle Theorem)
In △ABC, when AB=AC, prove that ∠B=∠C.
Answer: Let M be the intersection of the bisector of ∠A and BC.
In △ABM and △ACM / AB=AC (given) ① / ∠BAM=∠CAM (bisector of ∠A) ② / AM=AM (common side) ③ / From ①②③, △ABM≅△ACM by SAS. Therefore, ∠B=∠C because corresponding angles are equal.
In △ABC, when AB=AC, prove that ∠B=∠C.
Answer: Let M be the intersection of the bisector of ∠A and BC.
In △ABM and △ACM / AB=AC (given) ① / ∠BAM=∠CAM (bisector of ∠A) ② / AM=AM (common side) ③ / From ①②③, △ABM≅△ACM by SAS. Therefore, ∠B=∠C because corresponding angles are equal.
Key Points
- Draw an auxiliary line (the bisector of the vertex angle) to form two triangles
- Prove congruence by SAS, then reach the conclusion from the equality of corresponding angles
Practice Problem
- In △ABC, when AB=AC, prove that the bisector of the vertex angle intersects the base perpendicularly.
Answers and Explanations
- Answer: From the proof above, △ABM≅△ACM, so BM=CM and ∠AMB=∠AMC. Since ∠AMB+∠AMC=180° (a straight line), ∠AMB=90°. Therefore, AM⊥BC.