<p>Draw an isosceles triangle ABC with Side AB = Side AC. Draw a line from C to side AB and label that line CD. Now draw a line from B to side AC. Label that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE. </p>
<p>Find what angle EDC is by using geometry only and no trigonometry.</p>
<p>Don’t cheat and use protractors/rulers/all that stuff either! Go by pure geometric reasoning :D</p>
<p>Well difference here is that you actually can find this angle by pure geometric reasoning without using any weird trig functions. It's a challenging problem I think.</p>
<p>OK, you know why it's so hard? It's not mathematically possible. If those angles you gave us are true, then we'd end up with a triangle with angles of 20 - 30 - 130. Which we know is impossible because the sum of any two angles must be greater than the third angle. So this is a trick question.</p>
<p>Yeah I mean if two sides weren't greater than a third side then you'd have... well, a triangle with two lines that can never touch and that is such a depressing notion (lovers at a distance)!</p>