Question

State and prove basic proportinality theorem.using these theorem show that intercepts on a transversal by three equidistant parallel lines are always equal

Answer 1

Basic Proportionality Theorem:

Before we understand what Basic proportionality theorem has to say, we need to understand a few basics of similarity of triangles first. A brief brush up of basics of similarity is given below.

If two triangles are similar to each other then,

i)    Corresponding angles of both the triangles are equal andii)   Corresponding sides of both the triangles are in proportion to each other.

Thus two triangles ΔABC and ΔPQR are similar  if,

i)     ∠A=∠P, ∠B=∠Q and ∠C=∠Rii)    ABPQ = BCQR = ACPR

Let us now state the Basic Proportionality Theorem which is as follows:

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

area of ΔAPQarea of ΔPBQ = 12 × AP × QN12 × PB × QN = APPB

Similarly, area of ΔAPQarea of ΔQCP = 12 × AQ × PM12 × QC × PM = AQQC  ………..(2)

According to the property of triangles, the triangles drawn between the same parallel 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 for similar triangles as stated above. Thus, we can say that ∆ABC ~∆APQ.

The Mid Point theorem is a special case of the basic proportionality theorem.

According to mid-point theorem a line drawn joining the midpoints of the two sides of a triangle is parallel to the third side.

Consider a ∆ABC.

Figure 3: Mid-Point Theorem

We arrive at the following conclusions from the above theorem:

If P and Q are the mid points of AB and AC, thenPQ || BC. We can state this mathematically as follows:

If P and Q are points on AB and AC such that AP = PB = 12 (AB) and AQ = QC = 12 (AC), then PQ || BC.

Also the converse of mid -point theorem is also true which states that the line drawn through the mid-point of a side of a triangle which is parallel to another side, bisects the third side of the triangle.