A
(1) Basic proportionality theorem.
In A ABC, if seg PQ || seg AC
P
AP
Il
then
ОС
BO
BP
B
С C
Q
Fig. 1.25
Answers
Answer:
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.