$ABCD$ is a quadrilateral with $AB=BC$ and $BC||DA$ and $\angle ABD = x$, $\angle CBD=3x$, $\angle ADC=8x$. Then find the value of $x$. Let $ABCD$ is a quadrilateral with $AB=BC$ and $BC||DA$. Now join the diagonal $BD$. Given that $\angle ABD = x$ and $\angle CBD=3x$, also $\angle ADC=8x$.
Now find the value of $x$.
My Attempt:  Join AC. Then $\angle BAC = \angle BCA = \angle DAC = y.$
Also $\angle CBD=3x$ implies $\angle ADB = 3x.$ So, $\angle CDB = 5x$. Again calculating all we have $y=\pi - 2x.$ But I am failed to find the value of $x$.
Please Help me to find the value of $x$. Thanks in Advance.
 A: Continuing from the angle chasing you have done;
Let $AB=BC=a$ and $CD=b$ we observe $\angle  DCB=180-8x$
Now: area of triangle ABC=area of triangle BCD
or $$\frac{1}{2}a^2\sin 4x=\frac{1}{2}ab\sin8x$$
$$a\sin4x=b\sin 8x....(1)$$
Also by sine rule in triangle BDC:
$$b\sin 5x=a \sin 3x....(2)$$
$$\frac{\sin 3x}{\sin 4x}=\frac{\sin 5x}{\sin 8x}$$
can  you end it now ?
A: Here is a solution using angle chasing, the angle bisector theorem and similarity. First of all, there should be somewhere a picture:

Not all data from the picture is given, but let us construct the missing points and compute the not given marked angles.
(1) First of all, let $F$ be the point making $ABCF$ a rhombus. So $F$ is the reflection o $B$ w.r.t. $AC$
(2) Because $AF\|BC$, the point $D$ is on $AF$. We draw the diagonals in $ABCF$. The line $BD$ is the the angle bisector of the angle $\widehat{ABF}=2x=x+x$, since it separates from $\widehat{ABC}=4x=x+3x$ the pieces $x$ and $3x$.
(3) We are chasing some further angles now. We have four right angles in $O=AC\cap BF$ built by the diagonals of the rhombus, so $\widehat{BAC}$, $\widehat{BCA}$,
$\widehat{FAC}$, $\widehat{FCA}$ are each $90^\circ-2x$.
Then $\widehat{ADB}$, considered inside $\Delta ADB$ is $180^\circ-x-2(90^\circ-2x)=3x$. (And because we know $8x=\widehat{ADC}$, the remained $5x$ are dedicated to $\widehat{BDC}$.) Consider now the triangle $\Delta ADC$. Two angles are $90^\circ-2x$ and $8x$. So its angle in $C$ is $90^\circ-6x$.
(4) We obtain now the important angle
$$
\widehat{DCF}=
\widehat{ACF}-
\widehat{ACD}=
(90^\circ-2x)-
(90^\circ-6x)=4x\ .
$$
(5)
So the triangle $\Delta DCF$ is isosceles. Let $E'=M$ be the reflection of $E$ w.r.t. $AC$. Then we have
$$
CD=DF=EF=AE=AE'\overset{\color{red}?}=EE'\ , 
$$
and we want to show also the last equality marked with ${\color{red}?}$.
(6) Using the similarity $\Delta CEE'\sim\Delta CFB$ and the fact that $BE$ bisects $\widehat{CBF}$, we have:
$$
\frac {EE'}{EF} =
\frac {CE}{CF} =
\frac {CE}{BC} =
\frac {EF}{BF} \ ,
$$
so $EE'=EF$.
(7) This implies $\Delta AEE'$ equilateral. Its angle in $A$ is thus $60^\circ$, and half of it is $30^\circ=\widehat{CAE}=90^\circ-6x$, which determines
$$
\color{blue}{\boxed{\ x=10^\circ\ .}}
$$
(8) This is one direction, the condition is necessary. It is also sufficient. To see this in the same picture, draw first $ABCF$ as above with $x=10^\circ$, then draw the equilateral triangle $\Delta AEE'$, and possibly also its reflected image w.r.t. $BF$, then $CDF$ is by construction isosceles with two $40^\circ$ angles, and the computation in (5) with the question mark over an other equality sign shows that $BE$ bisects $\widehat{CBF}$.
$\square$
(9) Bonus: We know $x=10^\circ$.
Let $N,O,P$ be the intersections of $AC$ with the rays $BD$, $BF$, $BE$. Then the angles $\widehat{CPF}$ and $\widehat{CDF}$ are each $100^\circ$, so $CFDP$ cyclic, so
$\widehat{CDP}=
\widehat{CFP}=10^\circ$.
