Question

prove that if a line is drawn paralel to one side of a triangle to intersect the other two sides in disinct points then the other two sides are divided in the same ratio

Answer 1

In other words, We are asked proved the Triangle proportionality theorem,

Here ABC is a triangle, PQ is a line drawn parallel BC, So we need to prove that it divides the Sides AB, AC in the same ratio that is,


$\frac{AP}{PB} = \frac{AQ}{QC}$

Join PC, QB as shown in the figure.

In Δ APQ Draw perpendicular bisectors m, n.

Now,

Area of ΔAPQ = 1/2 * AP * QN.

Area of ΔPBQ = 1/2 * PB * QN.

Area of ΔAPQ = 1/2 * AQ * PM

Area of ΔPCQ = 1/2 * CQ * PM

We know that, Triangles between the same parallel lines and on the same base have equal area.

So, Area of ΔPBQ = Area of ΔPCQ.

Now, Area of ΔAPQ/ Area of ΔPBQ = AP/PB.

Area of ΔAPQ / Area of ΔPCQ ( = ΔPBQ) = AQ/CQ .

So , AP / PB = AQ/CQ .

Hence, We proved that a line drawn parallel to one side of a triangle divides the other two sides in the same ratio.