Question

In the trapezoid ABCD (

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

AP:PC and BN:CN

Answer 1

AP:PC = BN:CN = 3:5

Explanation:

The horizontal line MN divides any line between AB and CD into the same ratio as AM:MD, which is 3:5.

AP:PC = BN:CN = 3:5

_____

Since MP is parallel to AC, triangles AMP and ADC are similar with the scale factor AM:AD = 3:8. Likewise, triangles CNP and CBA are similar with the scale factor 5:8. These similarities and scale factors make ...

(AP+PC)/AP = 8/3

1 +PC/AP = 1 +5/3 . . . . . rewrite

AP/PC = 3/5 . . . . . . . . .subtract 1 and invert

__

(CN+BN)/CN = 8/5

1 +BN/CN = 1 +3/5 . . . . . rewrite

BN/CN = 3/5 . . . . . . . . . subtract 1