Example — carrying out the angle-bisector proof

Example — Carrying out the angle-bisector proof

In triangles $ABM$ and $ACM$ we have:

• $AB \cong AC$ (by hypothesis, isosceles triangle);
• $\widehat{BAM} \cong \widehat{CAM}$ (by hypothesis, $AM$ is the bisector);
• $AM$ is a common side.

By the first congruence criterion (side–angle–side): $\triangle ABM \cong \triangle ACM$. From this it follows that:

• $BM\cong CM$ $\implies$ $AM$ is a median;
• $\widehat{AMB}\cong\widehat{AMC}$; but the two angles are supplementary (they form the straight angle $BMC$), so each equals $\frac{\pi}{2}$ $\implies$ $AM$ is an altitude. $\blacksquare$

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Topics: Working method
Concepts: Write-up
Skills: Proving · Synthetic geometry