Question

In the trapezoid ABCD ( AB ∥ CD ) point M∈ AD , so that AM:MD=3:5. Line l ∥ AB and going through point M intersects diagonal AC and leg BC at points P and N respectively. Find: AC:PC

Answer 1

AC:PC=8:5

Explanation:

Draw a vertical line from D perpendicular to AB. Denote D' the intersection with I, and D'' intersection with AB. Consider the right triangle ADD". Realize that the lengths |DD"| and |DD'| are 3:5, which follows from triangles ADD" and MDD' being similar. This means that I bisects the trapezoid's height in a ratio 3:5.

Next, draw a vertical line perpendicular to AB, going through P. Denote P" the intersection of this line with AB, and denote P' the intersection of this line with DC. Observe that because lines CD || AB, the angles <BAC and <ACD are congruent. This implies triangles APP" and CPP' are similar with a side ratio 3:5, i.e., |AP|/|PC|=3/5. Therefore |AC|/|PC|=(|AP|+|PC|)/|PC|=|AP|/|PC|+1 = 8/5

The ratio AC:PC is 8:5