SOLUTION: Given rectangle ABCD, let P be the point outside ABCD that makes triangle CDP equilateral, and let Q be the point outside ABCD that makes triangle BCQ equilateral. Prove that trian

Question 147491: Given rectangle ABCD, let P be the point outside ABCD that makes triangle CDP equilateral, and let Q be the point outside ABCD that makes triangle BCQ equilateral. Prove that triangle APQ is also equilateral.

Would APQ be equilateral if ABCD was a parallelogram?

I happen to be very, very lost in this specific problem. I understand that it is equilateral, but unfortunately, I don't know how to prove it! Thanks!
Answer by vleith(2983) (Show Source): You can put this solution on YOUR website!
Draw the picture. Mark the two sets of congruent sides in the rectangle. Then layout the two equilateral triangles and mark the sides that are congruent with the two different sets in the rectangle. You end up with a rectangle with two triangles and only two different lengths for the various segments.
Remember that side-angle-side (SSA) is one means to establish congruent triangles.
Now:
The angles in the original rectangle are all 90 degree each.
The angles in equilateral triangles are 60 degrees each.
Triangles ADP, ABQ and CPQ are all congruent using SSA. How do we get that? Angle ADP is 90+60 = 150. Segment AD and DP as sides
Angle ABQ is also 90+60. Again BC and AB as sides
Angle PCQ = 360 - 60 - 90 - 60 = 150. Sides are CQ and CP
So we have 3 congruent triangles. Hence the thrid side of each is the same length. And the triangle composed of those three sides must be equilateral.
Not so for parallelogram
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