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 Î”ABC and Î”PQR are similar if,

- i) âˆ A=âˆ P, âˆ B=âˆ Q and âˆ C=âˆ 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 triangleÎ”ABC as shown in the given figure. In this triangle, we draw a line PQ parallel to the side BC of Î”ABC 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 Î”ABC 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 PMâŠ¥AC as shown in the given figure.

Now the area of âˆ†APQ = \( \frac12 \) Ã— AP Ã— QN (Since, area of a triangle= \( \frac12 \) Ã— Base Ã— Height)

Similarly, area of âˆ†PBQ= \( \frac12 \) Ã— PB Ã— QN

area of âˆ†APQ = \( \frac12 \) Ã— AQ Ã— PM

Also,area of âˆ†QCP = \( \frac12 \) Ã— QC Ã— PM â€¦â€¦â€¦â€¦ (1)

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

\( \frac {area ~of~ âˆ†APQ}{area~ of~ âˆ†PBQ}\) = \( \frac {\frac 12 ~Ã—~AP~Ã—~QN}{ \frac 12~Ã—~PB~Ã—~QN}\) = \( \frac {AP}{PB} \)

Similarly, \( \frac {area~ of~ âˆ†APQ}{area~ of~ âˆ†QCP}\) = \( \frac {\frac12~ Ã—~ AQ~Ã—~PM}{\frac 12 ~Ã—~QC~Ã—~PM}\) = \( \frac {AQ}{QC}\) â€¦â€¦â€¦..(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 âˆ†PBQ and QCP have the same area.

area of âˆ†PBQ = area of âˆ†QCP â€¦â€¦â€¦â€¦..(3)

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

\( \frac {AD}{PB}\) = \( \frac {AQ}{QC} \)

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