I came with this geometry problem and numerous lengthy solutions were proposed, so I thought there must be something missing in my solution.
The problem: Given that $\angle CAB=3x$, $\angle BCA=x$, and $\angle BDA=45$, determine the value of $x$.
And here is my sort of proof-without-words type of solution. (Thus $\angle x = 22.5$.)
Okay. Does adding this line clarify it? $\angle ABD=\beta$ and $\angle DBC=\alpha$. Then, $3x+\beta = 135$ and $\alpha + x=45$, thus $\beta=3\alpha$.
On
The key is to consider the point $X$ on $BC$ such that $AX=CX$. Then $\angle CAX = \angle XCA = x$, hence $$\angle BXA = \angle XCA + \angle CAX = x + x = 2x = 3x - x = \angle CAB - \angle CAX = \angle XAB.$$ Hence $AB=BX$.
Also, $$\angle XDB = 90^\circ - \angle BDA = 90^\circ - 45^\circ = 45^\circ = \angle BDA.$$
Using the well-known fact that the angle bisector and perpendicular bisector intersect at a point on the circumcircle, we get that $X,D,A,B$ lie on a circle. Hence $$2x = \angle BXA = \angle BDA = 45^\circ,$$ so $x = 22.5^\circ$.
The proof requires $\angle ACB = 90^0$.
Here is an indirect method of verifying that x is equal to $22.5^0$ (instead of giving a wild guess).
1) Create the semi-circle AXB (center = D, radius = DB) as shown.
2) Locate C’ on the circumference such that $\angle ADC’ = 45^0$.
3) Join BC’ such that $\angle DBC’ = y$. Then, $y = … = 22.5^0$.
It should be clear that, $z = … = 67.5^0$, which happens to be 3 times of y.
After mapping back with C’ = C, y = x, etc., we can say that x = 22.5 is one of the possible solution(s).
At a glance there is another location for C (which is $C_1$) provided $\theta = 3 \phi$. By moving $C_1$ along DC (and extended), geogebra shows such relation is not possible.