In △abc, point p∈ab so that ap: bp=1: 2. point r ∈ pb that pr: br=1: 2. point q∈bc

Ratio of PQC to ABC is 2:9.

Step-by-step explanation:

In ΔBRQ and ΔBPC

∠BQR = ∠BCP (given) and ∠B is common for both triangles, so from AAA similarity ΔBRQ and ΔBPC are similar.

⇒ BQ : QC = BR : RP = 2 : 1    →(1)

now draw perpendiculars from points A and P to BC line segment. Call the projected points as A' and P'.

It is clear that lines AA' and PP' are parallel. So in ΔBPP' and ΔBAA' we have AAA similarity with common angle at B.

⇒PP' : AA' = BP : BA = 2 : 3    →(2) (∵ AP : PB = 1 : 2)

area of ΔPQC = 0.5×PP'×QC

area of ΔABC = 0.5×AA'×BC

area of ΔPQC : area of ΔABC = (0.5×PP'×QC)/(0.5×AA'×BC)

                                               =(PP'/AA')×(QC/BC)

                                               =(2/3)×(1/3) (∵ from (1) and (2))

                                               =2/9.

∴ Ratio of PQC to ABC is 2:9

gracias :)

Step-by-step explanation: