Where Did E Come From?

Relevant wiki: Triangles

From given figure as, $BD = BE$ and so $DE || AC$ .

$\Rightarrow \angle BDE = 40^\circ \Rightarrow \angle a = 40^\circ , \angle c = 40^\circ$

$\Rightarrow \angle BDE = 40^\circ \Rightarrow \angle BED = 40^\circ \Rightarrow \angle b = 100^\circ , \angle e = 140^\circ$

$\Rightarrow \angle a + \angle b + \angle c + \angle e = 40^\circ + 100^\circ + 40^\circ + 140^\circ = \boxed{ 320^\circ}$

Looking at the triangle, it appears to be isosceles, which makes angle "a" congruent to angle "c", which both appear to be congruent to the angle measuring 40°. • a=40° • c=40° So this can give us angle "b". • b= 180-40-40= 100°

Now to solve for angle "e", we know this is an isosceles triangle, so we draw a straight line above angle "b", parallel to the base. We can name the angle between the line and the triangle angle "d". This is a similar situation to solving angle "a": we found an angle on the same straight line and tilted line. So, if you map it out, angle "e" is congruent to the sum of angle "d" and angle "b". • e=d+b To solve for "d", we can make a right angle, which we know is 90°. We can make this line going from the tip of the triangle to the midpoint of the base of the triangle, separating angle "b" in half. And we know that 100/2=50. But we're not done there. We are missing one angle in our complementary angle relationship. We can solve that by subtracting the angles in a right angle and the half we made of angle "b". • d=90-50=40°

Now we add "b" and "d" to get "e". • e=40+100=140°

Then we add everything.

a=40 b=100 c=40 e=140

a+b+c+e= 40+100+40+140= |320|

a + b + c =180 and e = 180 - 40 = 140... 180 + 140 = 320