If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the two sides in the same ratio. Let us now try to prove this statement.

Consider a triangleΔABC as shown in the given figure. In this triangle we draw a line PQ parallel to the side BC of ΔABC and intersecting the sides AB and AD in P and Q respectively.



Figure 1 Basic Proportionality Theorem

According the basic proportionality theorem as stated above, we need to prove:

ABPB=AQQC

Construction: Join the vertex B of ΔABC to Q and the vertex C to P to form the lines BQ and CP and then drop a perpendicular QN to the side AB and also draw PM⊥AC as shown in the given figure.



Figure 2 Basic Proportionality Theorem- Proof

Now the area of ∆APQ =12× AP × QN (Since, area of a triangle=12× Base × Height)

Similarly, area of ∆PBQ=12× PB × QN

area of ∆APQ =12× AQ × PM

Also,area of ∆QCP =12× QC × PM ………… (1)

Now, if we find the ratio of the area of triangles ∆APQand ∆PBQ, we have

areaofΔAPQareaofΔPBQ=12×AP×QN12×PB×QN=APPB

Similarly,areaofΔAPQareaofΔQCP=12×AQ×PM12×QC×PM=AQQC………..(2)

According to the property of triangles, the triangles drawn between the sameparallel linesand on the same base have equal areas.

Therefore we can say that ∆PBQ and QCP have the same area.

area of ∆PBQ = area of ∆QCP …………..(3)

Therefore, from the equations (1), (2) and (3) we can say that,

ADPB=AQQC

Also ∆ABC and ∆APQ fulfill the conditions forsimilar trianglesas stated above. Thus, we can say that ∆ABC ~∆APQ