A piece of paper has the precise shape of an equilateral triangle ABC which is creased along a line XY, with X on AB and Y on AC. so that A falls on some point D on BC.
(i) Are the triangles XBD and DCY always similar?
If so, prove it. If not, give a counterexample.
(ii) If AB = 15 and BD = 3, what is the length of the crease XY?
Geometry was never my strong suit but I think this works:
(i) Angles XBD, XDY, and DCY are all 60 degrees. Let angle ADY = n. Then angles AYD and BDX = (120 - n). So triangles XBD and DCY are always similar.
(ii) Note AB = BC. So if AB = 15 and BD = 3, then DC = 12 and CY = 3. Also angle DCY is 60 degrees, so we can use the law of cosines to compute the length of DY. We also note that DY = DX and angle XDY is 60 degrees, so XY = DY (triangle XDY is equilateral). Plugging into the formula gives XY = sqrt(117) ~= 10.82.
Edit: I think part i is ok but my answer to part ii is incorrect. I will revise it shortly.
Edited on March 5, 2014, 1:29 pm
Edited on March 5, 2014, 1:32 pm
Posted by tomarken
on 2014-03-05 13:23:18