The easiest hardest geometry question

Calculate some known angles: ACB = 180-(10+70)-(60+20) = 20° AEB = 180-60-(50+30) = 40°

Draw a line from point E parallel to AB, labeling the intersection with AC as a new point F and conclude: FCE ACB CEF = CBA = 50+30 = 80° FEB = 180-80 = 100° AEF = 100-40 = 60° CFE = CAB = 60+20 = 80° EFA = 180-80 = 100°

Draw a line FB labeling the intersection with AE as a new point G and conclude: AFE BEF AFB = BEA = 40° BFE = AEF = 60° FGE = 180-60-60 = 60° = AGB. ABG = 180-60-60 = 60°

Draw a line DG. Since AD=AB (leg of isosceles) and AG=AB (leg of equilateral), conclude: AD = AG. DAG is isosceles ADG = AGD = (180-20)/2 = 80° Since DGF = 180-80-60 = 40°, conclude: FDG (with two 40° angles) is isosceles, so DF = DG With EF = EG (legs of equilateral) and DE = DE (same line segment) conclude: DEF DEG by side-side-side rule DEF = DEG = x FEG = 60 = x+x

Answer: x = 30°

Consider the diagram.

$\angle BDA=180-20-60-50=50$

$\angle AEB=180-60-30-50=40$

Since $\angle ADB=\angle DBA=50$ , $\triangle ADB$ is isosceles with $AD=AB$ .

Let $AD=AB=1$ .

Apply sine law on $\triangle AEB$ .

$\dfrac{AE}{\sin 80}=\dfrac{1}{\sin 40}$ $\implies$ $AE \approx 1.532$

Appy cosine law on $\triangle ADE$ .

$(DE)^2=1^2+1.532^2-2(1)(1.532)(\cos 20)$ $\implies$ $DE \approx 0.684$

Apply sine law on $\triangle ADE$ .

$\dfrac{\sin x}{1}=\dfrac{\sin 20}{0.684}$

$x=\sin^{-1}0.5=$ $\boxed{30^\circ}$

Draw the figure and measure the angle. :P