Given a quadrilateral $ABCD$, $\angle DAC=3x$, $\angle CAB=4x$, $\angle BDC=2x$. Find the value of $x$

Question

This is a question I came across on Instagram today, and here's the diagram:

(Note: The image is NOT to scale) I attempted to solve it first by amending the quadrilateral in various ways, but each of those methods lead to a strange contradiction, likely because the diagram is not to scale.

I'm going to post my successful approach as an answer below, please let me know if my answer is correct or if there's something wrong with the method (the correct answer wasn't revealed, and if there are any other ways to approach this that I missed!)

Answer 1

$$\angle DCA=3x=\angle DAC\Longrightarrow DC=DA=a$$ Use sine law in $\triangle ABD$ $$\angle ADB=\pi-8x\Longrightarrow \frac{AB}{\sin(\pi-8x)}=\frac{a}{\sin x}\tag{1}$$ Use sine law in $\triangle BCD$ $$\frac{BC}{\sin(2x)}=\frac{a}{\sin x}\tag{2}$$ Use sine law in $\triangle ABC$ $$\angle ACB=\pi-6x\Longrightarrow \frac{AB}{\sin(\pi-6x)}=\frac{BC}{\sin(4x)}\tag{3}$$

Combine (1), (2), (3) give us

$$\begin{align}\sin(2x)\sin(6x)&=\sin(4x)\sin(8x)\\ \\ \Longrightarrow\sin(6x)&=2\cos(2x)\sin(8x)\\ \\ \Longrightarrow\sin(6x)&=\sin(8x+2x)+\sin(8x-2x)\\ \\ \Longrightarrow0&=\sin(10x)\\ \\ \Longrightarrow x&=k\cdot 18^\circ,~~~~k=1, 2, 3,... \end{align}$$

But eq.(1) tells us $x<22.5^\circ$, therefore,

$$x=18^\circ$$

Answer 2

This is my approach:

Note that $\angle AED=5x$. This means that $\angle ACD=3x$. Therefore, $AD=CD$. Let $F$ be a point on $BD$ such that $\angle BFC=\angle BAC=4x$. This implies that $\angle FCD=\angle FDC=2x$, $FD=FC$ and $\angle FCA=x$. This proves that quadrilateral $ABCF$ is cyclic, therefore $\angle FAC=x$ as well. This implies that $FD=FC=FA$, therefore $\angle FDA=2x$ as well.

Hence, $10x=180^\circ$, $x=18^\circ$

Answer 3

Another way to prove that $\;x=18\unicode{176}\,.$

Let $\,A’\,$ be the symmetric point of $\,A\,$ with respect to the segment $\,BD\,$ and let $\,H\,$ be the intersection of the segments $\,AA’\,$ and $\,BD\,.$
It results that $\,\angle DHA=90\unicode{176}\;$ and $\;\angle BHA=90\unicode{176}\,.$

There are three possible cases :

$1)\;\;H\;$ is on the right of $\,E\;;$
$2)\;\;H\;$ is on the left of $\,E\;;$
$3)\;\;H\;$ coincides with $\,E\;.$

In the first case, by applying Exterior Angle Theorem to the triangles $\,ABE\,$ and $\,CDE\,,\,$ we get that

$\angle DEA=5x\;\;$ and $\;\;\angle DCE=3x\;\;,$

consequently the triangle $\,DCA\,$ is isosceles and $\;DC\cong DA\,.$

On the other hand, since $\,A’\,$ is the symmetric point of $\,A\,$ with respect to $\,DB\,,\,$ it results that $\,DA\cong DA’\;,$

hence $\;DC\cong DA’\;$ and the triangle $\,DCA’\,$ is isosceles, but it is impossible because the angles $\,\angle DCA’\,$ and $\,\angle DA’C\,$ are not congruent, indeed

$\color{brown}{\angle DCA’}\mathbf{\,>\,}\color{brown}{\angle DCE}\mathbf{\,=\,}3x\quad$ and
$\color{brown}{\angle DA’C}\mathbf{\,=\,}\color{brown}{\angle A’BD}\mathbf{\,+\,}\color{brown}{\angle A’DB}\mathbf{\,<\,}x\mathbf{\,+\,}2x\mathbf{\,=\,}3x\;.$

So it is impossible that $\;H\;$ is on the right of $\,E\;.$
Analogously we can prove that it is impossible that $\;H\;$ is on the left of $\,E\;,$ hence the point $\,H\,$ coincides with $\,E\,$.

Consequently ,

$5x\mathbf{\,=\,}\color{brown}{\angle DEA}\mathbf{\,=\,}\color{brown}{\angle DHA}\mathbf{\,=\,}90\unicode{176}\;,$

$x=18\unicode{176}\,.$

Answer 4

In the attached figure, vertices $A',C'$ are obtained from a symmetrical mirroring over the line $DB$.

The $x$ value is obtained from triangle $\Delta A'BD$ which is congruent to $\Delta ABD$ where

$$ 180^{\circ}-(x+3x+4x)\le 2x $$

This happens to all possible configurations involving $A,A'$ and $C,C'$. Now if $C' = A$ and consequently $C=A'$ then

$$ 180^{\circ}-(x+3x+4x)= 2x \Rightarrow x = 18^{\circ} $$