# If $D\in [AB$ s.t. $AD=BC$ and $\angle{ADC}=\frac{3}{4}\cdot \angle{ABC}$ find $\angle {A}$.

Let $$\triangle {ABC}$$ s.t. $$AB=AC$$ and $$\angle{A}>90$$. If $$D\in [AB$$ s.t. $$AD=BC$$ and $$\angle{ADC}=\frac{3}{4}\cdot \angle{ABC}$$ find $$\angle {A}$$.

My idea: I denote $$\angle B=4x$$, then I apply "Sine theorem" in $$\triangle {ABC}$$ and $$\triangle {ADC}$$.

I obtain $$sin(5x)=2sin(3x)\cdot cos(4x)$$. Now I am stuck.

I try to construct and to solve it with an elementary construction but I didn't succeed. Can I apply here "The pants theorem"?

Let $$E$$ on $$BC$$ such that $$CE\cong AC\cong AB$$, and denote $$\angle ABC = \alpha$$.

1. $$\triangle BDE$$ is isosceles, therefore $$\angle BDE =\angle BED= \frac{\alpha}2$$.
2. Since $$\angle ADC = \frac34\alpha$$, we have $$\angle EDC = \frac14 \alpha$$.
3. Note that $$\angle DEC = 180^\circ -\frac12\alpha$$, so that $$\triangle DEC$$ is isoceles, too.
4. Construct then the rhombus $$DGCE$$ and connect $$A$$ with $$G$$. Observe that $$\triangle AGC$$ is equilateral, $$\triangle ADG$$ is isosceles, and $$DG\parallel BC$$.
5. We can write therefore $$\angle BAC$$ in two ways and obtain the equation $$180^\circ - 2\alpha = 60^\circ + \alpha.$$
6. Thus $$\alpha = 40^\circ$$ and $$\angle BAC = 100^\circ$$.

Continue with what you have and apply the trig identity $$2\cos a\sin b = \sin(a+b)-\sin(a-b)$$,

$$\sin5x=2\cos4x\sin3x=\sin7x-\sin x$$

Rearrange and apply the above identity one more time,

$$\sin x = \sin7x - \sin5x = 2\cos 6x\sin x$$

which yields $$\cos6x = \frac12$$. Thus, $$x=10^\circ$$, $$\angle B = 40^\circ$$ and $$\angle A = 180^\circ - 2\angle B = 100^\circ$$.