# proof of isosceles triangle?

How do you prove this isosceles triangle?

Given $AC$ = $BC$

Prove: $m\angle A=m\angle B$

I've gotten to the angle bisector and SAS(side- angle- side), and I believe there is one more step after that. I don't know what it is.

• Do you mean Angle $C =$ Angle $B$? – Henry Sep 2 '15 at 23:19
• You wrote "line AC is congruent to line BC" but in the diagram you marked AB and AC. – Akiva Weinberger Sep 2 '15 at 23:26
• sorry flipped the triangle around. – user242559 Sep 2 '15 at 23:31
• – Blue Sep 3 '15 at 0:20

Short answer: Once you have proven the two triangles congruent via SAS (or however you did it), you only need to select corresponding angles of the congruent triangles; those will be identical in measure.

Incidentally, the classic proof requires the construction of the angle bisector. Here's one way to do it without any additional lines:

• $AC = BC$ (given)
• $BC = AC$ (symmetry)
• $BC = CB$ (identity)
• $\triangle CAB \cong \triangle CBA$ (SSS)
• $m\angle CAB = m\angle CBA$ (corresponding angles of congruent triangles)

Attributed to Pappas, I believe.

• Actually the "classic" proof does not require the construction of an angle bisector. Euclid's proof (Book I, Proposition 5) involves extending both AB and AC by some arbitrary (but equal) amount and creating two distinct overlapping triangles, which share a common vertex angle at A, and hence are congruent by SAS. Then there is some additional detail work involving subtracting congruent parts from congruent angles. The construction is known as the "pons asinorum"; see en.wikipedia.org/wiki/Pons_asinorum. – mweiss Sep 2 '15 at 23:39
• Heh, sorry, I meant "classic" as in "classic modern textbook" (I guess that's an oxymoron). I have Elements at home; I'll have to take a look at that, thanks! – Brian Tung Sep 3 '15 at 1:04