Problem 4

Let

ABC

be a triangle with

AB = AC

. The angle bisectors of

\angle CAB

and

\angle ABC

meet the sides

BC

and

CA

D

and

E

, respectively. Let

K

be the incenter of triangle

ADC

. Suppose that

\angle BEK = 45^\circ

. Find all possible values of

\angle CAB

Step 3 of 5: Locate E via the angle bisector from B

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E=\left(\dfrac{\tan2x}{1+2\sin2x},\ \dfrac{2\sin2x}{1+2\sin2x}\right)

Detailed analysis

The base angles are $\angle ABC=\angle ACB=90^\circ-2x$ . By the angle bisector theorem applied to $BE$ in triangle $ABC$ , $\dfrac{AE}{EC}=\dfrac{AB}{BC}=\dfrac{AC}{BC}=\dfrac{\sec2x}{2\tan2x}=\dfrac{1}{2\sin2x}$ (using $AB=AC=\sec2x$ and $BC=2\tan2x$ ). Hence $E=A+\dfrac{1}{1+2\sin2x}(C-A)=\left(\dfrac{\tan2x}{1+2\sin2x},\ \dfrac{2\sin2x}{1+2\sin2x}\right)$ .