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.
[Math] Angle Bisectors in a Triangle
anglebisectioneuclidean-geometrygeometrytriangles
Related Solutions
Hope the following sketch can help.
Let $K$ be placed on the line $AC$ such that $A$ placed between $F$ and $K$.
Thus, $$\measuredangle CAD=\measuredangle DAB=\measuredangle BAK=60^{\circ},$$ which says that $AB$ is a bisector of $\angle DAK$.
But also $CE$ is a bisector of $\angle ACB$ and $$CE\cap AB=\{E\},$$ which gives that $DE$ is a bisector of $\angle ADB$.
By the same way we can show that $DF$ is a bisector of $\angle ADC,$
which says $$\measuredangle EDF=\frac{180^{\circ}}{2}=90^{\circ}.$$
I used the following facts.
Each point of an angle bisector is equidistant from the sides of the angle.
and
If a point is placed inside the angle and equidistant from the sides of the angle then this point is placed on the bisector of the angle.
Let $d(A,l)$ be the distance between a point $A$ and a line $l$.
Since $CE$ is a bisector of $\angle ACB$, we obtain: $$d(E,AC)=d(E,BC).$$ Since $AB$ is a bisector of $\angle DAK$, we obtain: $$d(E,AK)=d(E,AD).$$ But lines $AC$ and $AK$ they are the same.
Also, lines $DB$ and $BC$ they are the same.
Thus, $$d(E,AD)=d(E,AK)=d(E,AC)=d(E,BC)=d(E,DB),$$ which says $$d(E,AD)=d(E,DB)$$ and from here the ray $DE$ is a bisector of $\angle ADB$.
Best Answer
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$.