Question

State and Prove "Basic proportionally theorem"​

Answer 1

Answer:

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 by the same ratio .this is known as basic proportionality theorem .

Answer 2

Answer:

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.

Step-by-step explanation:

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

AP/PB = 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

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.