Question 1059721
Given: AB ≅ AC
  M is the midpoint of BC
Prove: AM bisects ∠BAC

{{{drawing(200,200,-.5,.5,-.5,2.5,triangle(-.3,0,0,0,0,2),triangle(.3,0,0,0,0,2),
locate(-.3,0,B), locate(.3,0,C),locate(0,2.2,A),locate(0,0,M)


)}}}
<pre><b>
AB &#8773; AC           |  Given
                  | 
&#916;ABC is isosceles | Definition of isosceles &#916;
                  |
&#8736;B &#8773; &#8736;C          | Base &#8736;s of an isosceles &#916; are &#8773;
                  | 
BM &#8773; CM           | Given that M is the midpoint of BC 
                  | 
&#916;ABM &#8773; &#916;ACM       | Side-angle-side 
                  |
&#8736;BAM &#8773; &#8736;CAM      | Corresponding parts of &#8773; &#916;s.
                  |
AM bisects &#8736;BAC   | It's two parts are &#8773;.

Edwin</pre></b>