An Angle Relationship In This Particular Triangle

1. Calculate some known angles: ACB = 180-(10+70)-(60+20) = 20° AEB = 180-70-(60+20) = 30°
2. Draw a line from point D parallel to AB, labeling the intersection with BC as a new point F and conclude: DCF ACB CFD = CBA = 60+20 = 80° DFB = 180-80 = 100° CDF = CAB = 70+10 = 80° ADF = 180-80 = 100° BDF = 180-100-20 = 60°
3. Draw a line FA labeling the intersection with DB as a new point G and conclude: ADF BFD AFD = BDF = 60° DGF = 180-60-60 = 60° = AGB GAB = 180-60-60 = 60° DFG (with all angles 60°) is equilateral AGB (with all angles 60°) is equilateral

4. CFA with two 20° angles is isosceles, so FC = FA

5. Draw a line CG, which bisects ACB and conclude: ACG CAE FC-CE = FA-AG = FE = FG FG = FD, so FE = FD

6. With two equal sides, DFE is isosceles and conclude: DEF = 30+x = (180-80)/2 = 50 Answer: x = 20°

$Let~AE~intersect~BD~at~P, ~~~and~~X=\angle~PED.\\ ~~~ \\ Apply~sum~ of~ angles~ of~ a~ \Delta~ is~ 180^o.\\ \Delta~PAB,~~~\angle~BPA=50........(0)\\ \Delta~DAB,~~~\angle~BDA=40.\\ \Delta~EAB,~~~\angle~BEA=30.\\ (0)~\implies~~~~\angle~EPB=130.~~~~~\therefore~internal~\color{#3D99F6} {\angle~EDP=130-X.} \\ ~~~~~\\ ~~~ \\ Apply~Sine~law.\\ \Delta~BDA,~~~~AB=AD*\dfrac{Sin40}{Sin60}................(1)\\ \Delta~BEA,~~~~AB=BE*\dfrac{Sin30}{Sin70}................(2)\\ \Delta~EDA,~~~~ED=AD*\dfrac{Sin10}{SinX}................(3)\\ \Delta~EDB,~~~~ED=BE*\dfrac{Sin20}{Sin(130-X)}.........(4)\\ ~~~~\\ ~~~ \\ From~(1)~and~(2)\dfrac{AD} {BE}=\dfrac{\frac{Sin30}{Sin70}} {\frac{Sin40}{Sin60}}.\\ From~(3)~and~(4)\dfrac{AD} {BE}=\dfrac{\frac{Sin20}{Sin(130-X)}} {\frac{Sin10}{SinX}}\\ ~~~~\\ ~~~ \\ \therefore~~\dfrac{Sin40*Sin70*Sin20}{Sin60*Sin30*Sin10}=\dfrac{Sin(130-X)}{SinX}\\ =Sin130*CotX-Cos130.\\ ~~~~\\ ~~~ \\ \therefore~CotX=\dfrac{Sin40*Sin70*Sin20}{Sin60*Sin30*Sin10*Sin130}+Cot130 \\ X=Cot^{-1} \left (\dfrac{Sin40*Sin70*Sin20}{Sin60*Sin30*Sin10*Sin130}+Cot130 \right )=\Large~~\color{#D61F06}{20^o}.$

simple

From properties of triangle CE/sin10 = CA/sin150 in CAE CD/sin20 = CB/sin 140 in isosceles triangle ABC, CA=CB So CE/CD is available from earlier two relations. So sinCED/sinCDE is available. Also CED+CDE = 160 as DCE = 20 From the above CED comes as 130 So x=150-130 = 20

Draw the circumcircle of $\triangle CDE$ . Then prove that $\overline{AE}$ is tangent to the circumcircle and we have $x = \angle C = 40^\circ$ , by alternate segment theorem.

Use sine law in triangles: ADE: (sin x)/AD = (sin 10)/DE, ABD: (sin 60)/AD = (sin 40)/AB, and ABE: (sin 30)/AB = (sin 40)/DE. Multipy these equations together to cancel out AD, DE, and AB and derive the equation: sin x = sin 60 sin 10 sin 30 /(sin 20 sin 40), which computes correctly to the answer.