Quadrilateral $ABCD$ , $\angle ABD = 17^{\circ}, \angle DBC = 34^{\circ}, \angle ACB = 43^{\circ}, \angle ADB = 13^{\circ}$, Find $\angle BDC$.
[Math] Find $\angle BDC$
geometry
Related Solutions
I have four different and one general solution to this problem, I hope it will help you.
Well, Geogebra says it is $\approx 77,34^{\circ}$, so good luck...
Actually, Ceva might really help:
$${\sin 80\over \sin 40}{\sin(70-x)\over \sin x}{\sin 20\over \sin90} = 1$$
After some manipulation we get $$\cot x = \tan 20+{2\over \cos 10}\implies x =... $$
Best Answer
Nice problem!
Let $X$ be a point symmetric to $A$ with respect to $BD$. Let $Y$ be a point symmetric to $A$ with respect to $BX$. Let $Z$ be a point symmetric to $A$ with respect to $DX$.
Then $DA=DX=DZ$, $BA=BX=BY$, and $AX=XY=XZ$. Angle chasing gives $$\angle ZXY=360^\circ - \angle AXZ - \angle YXA = 360^\circ - 2\angle AXD - 2\angle BXA = \angle XDA + \angle ABX = 2\angle BDA + 2 \angle ABD = 2\cdot 13^\circ+2\cdot 17^\circ = 60^\circ$$ which along with $XZ=XY$ implies that $XYZ$ is equilateral. Thus $$\angle BZA = \angle BZX + \angle XZA = 30^\circ + 13^\circ = 43^\circ.$$ We also have $$\angle DBZ = \angle DBX + \angle XBZ = 2\angle ABD = 34^\circ.$$ This means that $Z=C$. Therefore $$\angle CDB = \angle ZDB = 3\angle BDA = 39^\circ.$$
Below is a trigonometric solution. Let $\angle CDB = x$. We use Snellius' theorem thrice: \begin{align} \frac{AC}{DC} & = \frac{\sin(13^\circ + x)}{\sin 64^\circ}, \\ \frac{DC}{CB} & = \frac{\sin 34^\circ}{\sin x}, \\ \frac{CB}{AC} & = \frac{\sin 86^\circ}{\sin 51^\circ}. \end{align} Multiplying yields $$1=\frac{\sin(13^\circ + x)\sin 34^\circ \sin 86^\circ}{\sin 64^\circ \sin x \sin 51^\circ}$$ so $$\sin(13^\circ + x)\sin 34^\circ \sin 86^\circ = \sin 64^\circ \sin x \sin 51^\circ.$$ Using $2\sin A \sin B = \cos(A-B) - \cos(A+B)$ twice we get $$\sin(13^\circ + x)\left(\cos 52^\circ - \cos 120^\circ \right) = \sin x \left(\cos 13^\circ + \cos 65^\circ\right).$$
Since $\cos 120^\circ = -\frac 12$, we have $$\sin(13^\circ + x) \cos 52^\circ + \frac 12 \sin(13^\circ + x) = \sin x \cos 13^\circ + \sin x \cos 65^\circ.$$
Multiplying by two and using $2\sin A \cos B = \sin(A+B) + \sin(A-B)$ we infer
$$\sin(65^\circ + x) + \sin(x-39^\circ) + \sin(13^\circ + x) = \sin(x+13^\circ) + \sin(x - 13^\circ) + \sin(x+65^\circ) + \sin(x-65^\circ).$$
Therefore $$\sin(x - 39^\circ) = \sin(x-13^\circ) + \sin(x-65^\circ).$$
We use now $\sin A + \sin B = 2 \sin \frac{A+B}2 \cos\frac{A-B}2$:
$$\sin(x-39^\circ) = 2\sin(x-39^\circ)\cos 26^\circ$$
or
$$\sin(x-39^\circ)(1-2\cos 26^\circ)=0.$$
Since $\cos 26^\circ \neq \frac 12$, we have $\sin(x-39^\circ)=0$ and so $x=39^\circ$.