Post Number:#2 by ahmedittihad » Thu Jun 22, 2017 4:53 am
Ugly problem tbh.
Let us assign $\angle BAC = 4a$. Then quick angle chases give us, $\angle CAK = a $, $\angle DCI = 45-a $, $\angle CEK = 3a $.
We can apply sine law on $\triangle IEK $ and $\triangle CEK $ to get $ \frac{IK}{KC} = \frac{sin(45)*sin (45-a)}{sin(3a)*sin(90-2a)}$ ........(relation 1).
By angle bisect or theorem and sine law again, $\frac{IK}{KC} = \frac {ID}{CD} = \frac{sin(45-a)}{sin(45+a)}$ ........(relation 2).
Using the two relations together, we get that $ sin(45)*sin(45+a) = sin(3a)*sin(90-2a)$.
Now we use the product-sum rule and get $cos(a) = cos(5a-90)$.
Which implies that $cos(5a-90) - cos(a) = 0$.
Applying the product-sum rule again yields $cos(5a-90) - cos(a) = 0 = 2 sin(3a-45)*sin(2x-45)$.
We know that $0 \leq a <45$. So the two solutions are $45/2$° and $15$°.
Frankly, my dear, I don't give a damn.