Prove btp theorem....
Answers
Basic Proportionality Theorem Proof
Statement:
If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion
Given:
In triangle ΔABC, 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 AC in P and Q, respectively
To prove:
AP/PB =AQ/QC
Proof
Now the area of ∆APQ = 1/2 × AP × QN (Since, area of a triangle= 1/2× Base × Height)
Similarly, area of ∆PBQ= 1/2 × PB × QN
area of ∆APQ = 1/2 × AQ × PM
Also,area of ∆QCP = 1/2 × 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 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,
AP/PB = AQ/QC
Also, ∆ABC and ∆APQ fulfil 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 an ∆ABC.
Mid-point Theorem
Conclusion
If P and Q are the mid-points of AB and AC, then PQ || BC.
If P and Q are points on AB and AC such that AP = PB = 1/2 (AB) and AQ = QC = 1/2 (AC), then PQ || BC
Hence, the basic proportionality theorem is proved.
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 in distinct points, then the other two sides are divided in the same ratio.
- Hence, the basic proportionality theorem is proved.
- Hence, proved