This is the picture, and we are aiming for the angle $x$
It's easy to see that $\angle DGA = \angle CGB = 100°$, $\angle CGD = \angle AGB = 80°$, $\angle CBG = 50°$, but now i'm missing $\angle GBA = ? $ and $\angle GAB = x$
Any hints?
Find the angle in a quadrilateral
euclidean-geometrygeometryquadrilateral
Related Solutions
This problem is a particular case of a family of problems with broadly the same solution, so I will post this more general solution and then discuss particular instances of it.
Problem. $\angle BAC=3\angle CAD$; $\angle CBD=30^\circ$; $AB=AD$. What is $\angle DCA$?
Solution. Let $\alpha=\angle CAD$. $\triangle BDA$ is isosceles on base $BD$. Therefore $\angle DBA=\angle ADB=90^\circ-2\alpha$ and $\angle CBA=120^\circ-2\alpha$.
Let $E$ be on $BC$ such that $AE=AB$. Then $\triangle BEA$ is isosceles on base $BE$. Therefore $\angle AEB=\angle EBA=120^\circ-2\alpha$, so $\angle BAE=4\alpha-60^\circ$, so $\angle EAD=60^\circ$.
Therefore $\triangle AED$ is equilateral, so $\angle EAC=60^\circ-\alpha=\angle ACE$, so $\triangle CAE$ is isosceles on base $CA$, i.e. $CE=AE=DE$, so $\triangle CDE$ is isosceles on base $CD$. $\angle CED=2\alpha$, so $\angle DCE=90^\circ-\alpha$, so $\angle DCA=30^\circ$, which solves the problem. Note that $\angle DCA$ is independent of $\alpha$.
To adapt this to the current problem, relabel from $ABCD$ to $BCDA$ and specify $\alpha=19^\circ$.
If $\alpha$ is specified as $20^\circ$, and $\angle DBA$ as $50^\circ$, then the problem is [Langley]. $AB=AD$ is easily seen, and the proof proceeds as above. The above proof, but with angles as specified in Langley's problem, is due to J. W. Mercer.
If $\alpha$ is specified as $16^\circ$, then the problem is that at gogeometry. The point-lettering is the same, but the diagram is flipped.
[Langley] Langley, E. M. "Problem 644." Mathematical Gazette, 11: 173, 1922, according to David Darling
Let $AD\cap BC=\{E\}$.
Thus, since $$\frac{AE}{CE}\cdot\frac{CE}{DE}\cdot\frac{DE}{BE}\cdot\frac{BE}{AE}=1$$ we obtain: $$\frac{\sin50^{\circ}}{\sin20^{\circ}}\cdot\frac{\sin60^{\circ}}{\sin50^{\circ}}\cdot\frac{\sin(70^{\circ}-x)}{\sin{x}}\cdot\frac{\sin80^{\circ}}{\sin30^{\circ}}=1$$ or $$\frac{\sin(70^{\circ}-x)}{\sin{x}}=\frac{\sin20^{\circ}\sin30^{\circ}}{\sin60^{\circ}\sin80^{\circ}}.$$ But it's obvious that $x$ is an acute angle and $\frac{\sin(70^{\circ}-x)}{\sin{x}}$ decreases.
Id est, it's enough to prove that: $$\frac{\sin10^{\circ}}{\sin60^{\circ}}=\frac{\sin20^{\circ}\sin30^{\circ}}{\sin60^{\circ}\sin80^{\circ}}$$ or $$2\sin10^{\circ}\sin80^{\circ}=\sin20^{\circ},$$ which is obvious.
Best Answer
By law of sines we obtain: $$\frac{BG}{GD}=\frac{\frac{BG}{AG}}{\frac{GD}{AG}}=\frac{\sin{x}}{\sin(100^{\circ}-x)}.$$ In another hand, $$\frac{BG}{GD}=\frac{\frac{BG}{CG}}{\frac{GD}{CG}}=\frac{\frac{\sin30^{\circ}}{\sin50^{\circ}}}{\frac{\sin70^{\circ}}{\sin30^{\circ}}}=\frac{1}{4\sin50^{\circ}\sin70^{\circ}},$$ which gives $$\frac{\sin{x}}{\sin(100^{\circ}-x)}=\frac{1}{4\sin50^{\circ}\sin70^{\circ}}$$ or $$\sin100^{\circ}\cot{x}-\cos100^{\circ}=4\sin50^{\circ}\sin70^{\circ}$$ or $$\cot{x}=\frac{\cos100^{\circ}+4\sin50^{\circ}\sin70^{\circ}}{\sin100^{\circ}},$$ which gives $x=20^{\circ}.$
Indeed, $$\cot{x}-\cot20^{\circ}=\frac{\cos100^{\circ}+4\sin50^{\circ}\sin70^{\circ}}{\sin100^{\circ}}-\cot20^{\circ}=$$ $$=\frac{\cos100^{\circ}+2(\cos20^{\circ}-\cos120^{\circ})}{\cos10^{\circ}}-\frac{\cos20^{\circ}}{\sin20^{\circ}}=$$ $$=\frac{2\sin10^{\circ}(\cos100^{\circ}+2\cos20^{\circ}+1)-\cos20^{\circ}}{\sin20^{\circ}}=$$ $$=\frac{2\sin10^{\circ}(\cos40^{\circ}+\cos20^{\circ}+1)-\cos20^{\circ}}{\sin20^{\circ}}=$$ $$=\frac{\sin50^{\circ}-\sin30^{\circ}+\sin30^{\circ}-\sin10^{\circ}+2\sin10^{\circ}-\cos20^{\circ}}{\sin20^{\circ}}=$$ $$=\frac{\sin50^{\circ}+\sin10^{\circ}-\cos20^{\circ}}{\sin20^{\circ}}=0$$ and we are done!