euclidean-geometrygeometrytriangles
I have to find the value of $x$ and I did it and found out $x=40^0$. I used the cot theorem to find it. but without trigonometry is there a purely geometric proof? every single time I tried I was stopped at $x+F\hat BD=80^0$. Any hint or a proof would be appreciated.
P.S this is different from Langley’s Adventitious Angles problem since $CE \neq AC$
I saw some basic constructions led me to have some cyclic quadrilaterals. but they also did not gave me the answer
My favorite interpretation comes from the following claim / 'obvious fact'
Claim: Amongst all smooth curves with a fixed perimeter $P$, the area $A$ satisfies the inequality $ 4 \pi A \leq P^2$. Equality holds when the curve is a circle.
Lemma: If the 4 given sides satisfy the cyclic inequalities $a < b+c+d$, then a quadrilateral with sides $a, b, c, d$ exists. Moreover, a cyclic quadrilateral with sides $a, b, c, d$ exists. A sketch of this is given at the end.
This Lemma merely serves to let us know when the question makes geometric sense. Now, let $ABCD$ be a cyclic quad with the desired side lengths, and consider it's circumcircle. Let it have area $X$. Fix the circular arcs along with the intermediate areas between the circular arcs and the quadrilateral. Let these arcs have perimeter $P$, and the intermediate area have total area $I$.
Take any deformation $A^*B^*C^*D^*$, where the arcs and the areas are moved under translation and rotation. This gives a figure with perimeter $P$, and area $I+X^*$, where $X^*$ is the area of the deformed $A^*B^*C^*D^*$. By the initial claim, $I+X^* \leq I+X$. Hence we are done.$ _ \square$
Sketch Proof of Lemma: The first part follows from the triangle inequality. The second part follows by taking a continuous deformation of the quadrilateral, showing that opposite angles must equal $180^\circ$ at some point in time.
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
Best Answer
In this picture, there are some things that have to be further explained:
First of all, we take the symmetry of $\Delta BEC$. Then we draw a line from $B$ to $B'$ in order to have an equilateral triangle $\Delta BEB'$. Then we have the equality $|EB| = |BB'| = |EB'| = |AB|$. Here, since $\angle EBB' = 60^\circ$ and $|AB| = |BB'|$, we can conclude that $A$, $F$, $D$ and $B'$ are linear and $\angle AB'B = 20^\circ$. Then by noticing the fact that $EO$ is median of the equilateral triangle, it is median of $\Delta BDB'$ as well, which implies that $\angle DBB' = 20^\circ$. Therefore $x = 40^\circ$.