Adventitious Quadrangle 4

hint:

1. reflect $D$ about $BC$ .

2. show that $ADD'$ is a straight line.

3. show that $ABD'C$ is a cyclic quadrilateral by inscribed angle theorem.

4. show that $x=30^{\circ}$ .

What I like about this problem is how there are many different ways to solve it. First, find $m\angle D$ , $m\angle E$ , and $m\angle F$ :

$D=180-(40+20)=120^{\circ}$ $E=180-(30+20)=130^{\circ}$ $F=360-(D+E)=360-(120+130)=110^{\circ}$

Since we have labeled $F$ , we now have a formula for $x$ : $x+F+(A-20)=180$ Then, draw a perpendicular bisector from $B$ to $AC$ :

Note that since $m\angle ABD = 30^{\circ}$ , $\triangle ABP$ is a 30-60-90 triangle. Therefore, $m\angle A=60^{\circ}$ .

Finally,

$x+F+(A-20)=180$ $x=180-110-40=\boxed{30^{\circ}}$

From trigonometric version of Ceva's theorem, we know that:

$\frac{sin\angle DBA}{sin\angle CBD} \cdot\frac{sin\angle CAD}{sin\angle DAB}\cdot\frac{sin\angle DCB}{sin\angle ACD}=1$

$\Rightarrow\frac{sin30} {sin40} \cdot \frac{sin(70-x)}{sin20}\cdot \frac{sin20} {sin x} =1$

Now it's only calculations:

$\frac{sin30} {sin40} \cdot \frac{sin70cosx-cos70sinx} {sinx} =1$

$\frac{sin70-cos70tanx}{tanx}=\frac{sin40} {sin30}$

$tanx=\frac{sin70}{\frac{sin40} {sin30} +cos70} \Rightarrow x=\boxed{30}$

Angle BDC=120°, Angle ADB=130° so Angle ADC=110°. Point D is the ortho-centre of ∆ABC as Angle ADC=180°-70= 110°. So angle BDA=180°-Angle C. Substituting we get Angle ACB as 50°. So x=30°