#### notes

**Theorem**: If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.

**Given**: In D ABC line l || line BC and line l intersects AB and AC in point P and Q respectively

**To prove**: `"AP"/"PB"="AQ"/"QC"`

**Construction**: Draw seg PC and seg BQ

Proof: Δ APQ and Δ PQB have equal heights.

`therefore (A(triangle APQ))/(A(triangle PQB))="AP"/"PB"` .....................(I) (areas proportionate to bases)

`therefore (A(triangle APQ))/(A(triangle PQC))="AQ"/"QC"`................... (II) (areas proportionate to bases)

seg PQ is a common to base of Δ PQB and ΔPQC. seg PQ || seg BC,

hence Δ PQB and Δ PQC have equal heights.

`A(triangle PQB)=A(triangle PQC)` ................(III)

`(A(triangle APQ))/(A(triangle PQB))=(A(triangle APQ))/(A(triangle PQC))` .......... [from (I), (II) and (III)]

`therefore "AP"/"PB"="AQ"/"QC"` .......... [from (I) and (II)]