Hardest Easy Geometry Problem

with 'chasing 180' i reached 3 equations with 4 variables, im trying to solve it right now
i will try after with his hint 2, drawing lines inside the triangle

Solution
LETS SOLVE FIRST ONE AND SECOND ONE GOES SIMILAR

.... I Guess



1. LET Angle(DEC) = X & Angle(AEB) = Y

So X+A+Y = 180
A+Y = 180 - X .......... EQ1


Now Angle(DCE) = 180 - (10+70) - (20+60)
Angle(DCE) = 180 - 160 = 20

Also A+X+10+Angle(DCE)=180 ... For Triangle(EAC)
A+X = 170 - Angle(DCE)
A+X = 170 - 20 = 150 ........ EQ2
X = 150 - A



Using EQ1

A + Y = 180 - 150 + A = 30 + A
Cancelling A from both Sides
Y = 30 ......... Ans(1)



Assume intersection point of line DB and line AE as Z

Angle(DZE) = Angle(AZB) ..... opposite angles
Angle(DZE) = Angle(AZB) = 180 - (70+60) = 50 ....... Triangle(AZB)
Let Angle(EDB) = P
A + 50 + P = 180 ...................Triangle(DZE)
P = (130 - A) ........... EQ3


Also using the theory of External Angles
For Triangle(DEB) for External Angle X
20+P = 180 - X
20+130-A=180-X ... FROM EQ3

So A-X = -30
Also A+X = 150 ... FROM EQ2


Solving Simultaneously .... A = 60 .... Ans 2

So X = 180-(A+Y)=180-(60+30)=90 ... Ans 3