Triangle ABC is isosceles with AB=AC. Point D is on side AB such that angle BCD is 70 degrees. Point E is on side AC such that angle EBC is 60 degrees. Angle ABE equals 20 degrees, and angle DCE equals 10 degrees.

Find angle EDC. Justify your answer.

(In reply to by )

Sorry I messed up and did more than was needed. I'll try again.

First , I draw the problem. An isosceles triangle,  vertex A at the top, B bottom right, C bottom left.
Put a point E on the left side, midway up between C and A.
Put a point D on the right side, midway up between B and A.
Draw a line between E and D.
Draw a line from E to B.
Draw a line from D to C.

I get a triangle with an "hour-glass" inside, or a black-widow mark. :-)  I marked the center point as F.

Now I put the angle values in that are given.

BCD = 70
EBC =  60
ABE = 20
DCE = 10.

I don't care that things are not drawn to scale as long as the numbers add up.  Inside angles of a triangle must add up to 180. Once you have 2 of the three angles of a given triangle, you know the third.  Also, when two lines intersect, the 2 angles on one side must add to 180. If you know one, you know the other.

Starting at the bottom of the drawing, start filling in the angle values from the ones you already know.

Angle A = 180 - C - B = 180 - 80 - 80 = 20.
Complete triangle FBC = 70 + 60 + ? & get 50 for the angle below F.
Work your way around the X formed at F. Get 130, 50, 130, 50.
Now that triangle on the left, EFC has values 10 & 130, so the angle below E is 40.

We now have all of the angles except  for the two at point E, and two at point D.

The two at E have to sum to 140 = 180 - 40.
The two at D have to sum to 150 = 180 - 30.

If you have followed me this far, you see the upside down kite I was refering to in my first post.  We know the top angle = 20 at A. The bottom angle = 50 at F. (This is where I was wrong in my first post. I had 70).

The two angles at the left have to sum to 140. The two at the right have to sum to 150, and one of these is the angle EDC, which we are looking for.

So, here we are. Just looking at the top two triangles, and plugging different values in to maintain the 180 rule, we can get many different values for angle EDC, all of which are valid.

No unique solution.

Posted by bob909 on 2004-11-04 08:13:37