Angle Bisectors in a Triangle My son got this problem in geometry and was stumped. He asked me and I am stumped too. Here is the problem:
In triangle ABC, m∠ACB = 42°. The angle bisectors AD and BE intersect at point O so that AE + OE = AB. Find m∠A and m∠B.  
 A: Say $F$ is on $AB$ so that $AF = AE$. Then $\triangle AEO \cong \triangle AFO $ (sas). So $$FO = EO = FB$$
Thus $\triangle BOF$ is isosceles. So $$\angle CEO = 180- \angle ABC$$ So $$180 -{1\over 2} \angle ABC +42 = 180 \Longrightarrow \angle ABC = 84$$
and  $ \angle CAB = 54$.
A: Hint: First you calculate $\angle AOB= 180^{\circ}-\dfrac{A}{2}-\dfrac{B}{2}= 180^{\circ}- \dfrac{1}{2}\left(180^{\circ}-42^{\circ}\right)=111^{\circ}$. Next use the property of bisector in $\triangle AEB$, you have:$\dfrac{OE}{OB} = \dfrac{AE}{AB}= \dfrac{OE+AE}{OB+AB}= \dfrac{AB}{OB+AB}\implies \dfrac{OE}{AB}= \dfrac{OB}{OB+AB}\implies \dfrac{OB}{OB+AB}+\dfrac{AE}{AB} = \dfrac{OE}{AB}+\dfrac{AE}{AB}=1\implies \dfrac{\dfrac{OB}{AB}}{\dfrac{OB}{AB}+1}+\dfrac{AE}{AB}=1\implies \dfrac{\sin (A/2)}{\sin 111^{\circ}+\sin(A/2)}+\dfrac{\sin (B/2)}{\sin(42^{\circ}+B/2)}=1\implies \dfrac{\sin (A/2)}{\sin 111^{\circ}+\sin (A/2)}+\dfrac{\sin(69^{\circ} - A/2)}{\sin(111^{\circ} - A/2)}=1$. Observe that: $\sin(111^{\circ} - A/2) = \sin (69^{\circ} + A/2)$. I can continue to the end but leave it for your to finish it. You can solve for $\sin (A/2)$ and then $A/2$ and double it to get $A$, then $B$.
