fluke wrote:
Say,
ABCD is the rectangle. AC and BD intersect at O. Draw a perpendicular from O on AB such that it intersects at P. Now,
ABC ~ APO
AC/AO=BC/PO
AC=2AO(Diagonals bisect each other)
BC/PO=2
PO=1/2*BC
But we must prove that the triangles, ABC, APO are similar using the SAS theorem. (2 sides in proportion and the angle between them will be equal)
1.AP and AB are in proportion (because the altitude, AP (of APO triangle) will also be median, since AOB is isosceles triangle. It is isosceles because the diagonals bisect each other and AO=BO)
2.AO and AC are in proportion (since in rectangle the diagonals bisect each other)
3.they share the included angle CAB=OAP
Therefore the sides of triangles of APO:ABC will be 1:2, therefore the ratio of PO:BC will be 1:2 also.
Or trying to prove similarity, we could say that they have 2 angles congruent, therefore they are similar.
Correct?
Thanks so much fluke for your input!