A rectangular piece of paper PQRS has sides Pq = 12 cm and PS = 13 cm. The pt O is the midpt of PQ. The pts K and M are to be chosen on OQ and PS respectively so that when the paper is folded along KM, the corner that was at P lands on the edge QR at L. Let OK = x cm and LM = y cm.
For a part of the question, I have to show that triangles QKL and NLM are similar. I know that the reason is: equiangular. However, I cannot seem to prove that a second set of angles are equal in these two triangles. Could someone please help me?
April 13th 2009, 08:06 AM
If we look at the angle KLQ, we see it sits on a straight line along with a right angle. So this angle is determined by subtracting MLN and 90 from 180 (angles on a straight line)
In triangle MLN, the angle LMN is determined in exactly the same way (this time since all angles in a triangle sum to 180), therefor KLQ and LMN are the same. Since both triangles also have a right angle in them, it follows that the third angle in each is the same