# Basic Proportionality Theorem & Similar Triangles

Before we understand what Basic proportionality theorem states, 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 and
• ii) 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=∠R
• ii) $\frac {AB}{PQ}$ = $\frac {BC}{QR}$ = $\frac {AC}{PR}$

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 the 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.

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

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

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.

Now the area of ∆APQ = $\frac12$ × AP × QN (Since, area of a triangle= $\frac12$ × Base × Height)

Similarly, area of ∆PBQ= $\frac12$ × PB × QN

area of ∆APQ = $\frac12$ × AQ × PM

Also,area of ∆QCP = $\frac12$ × QC × PM ………… (1)

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

$\frac {area ~of~ ∆APQ}{area~ of~ ∆PBQ}$ = $\frac {\frac 12 ~×~AP~×~QN}{ \frac 12~×~PB~×~QN}$ = $\frac {AP}{PB}$

Similarly, $\frac {area~ of~ ∆APQ}{area~ of~ ∆QCP}$ = $\frac {\frac12~ ×~ AQ~×~PM}{\frac 12 ~×~QC~×~PM}$ = $\frac {AQ}{QC}$ ………..(2)

According to the property of triangles, the triangles drawn between the same parallel lines and 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,

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

Also, ∆ABC and ∆APQ fulfill the conditions for similar triangles as stated above. Thus, we can say that ∆ABC ~∆APQ.

The MidPoint 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.

We arrive at the following conclusions from the above theorem:

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

If P and Q are points on AB and AC such that AP = PB = $\frac 12$ (AB) and AQ = QC = $\frac 12$ (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.

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