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

Question

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"?

Answer 1

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 $\angle BAC = 100^\circ$.

Answer 2

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 $\angle A = 180^\circ - 2\angle B = 100^\circ$.