Match the following column I with II Let P be an interior point of the triangle ABC and  the lines AP,BP and CP when produced meet the  opposite sides in D,E and F respectively then PDAD+PEBE+PFCF is equal to

(P) We have, rr1=4Rsin⁡A2sin⁡B2sin⁡C24Rsin⁡A2cos⁡B2cos⁡C2=tan⁡B2tan⁡C2=14∴tan⁡A2tan⁡B2+tan⁡C2=1−tan⁡B2tan⁡C2=1−14=34 (Q) We have, r1r2r31/3≥31r1+1r2+1r3=3r∴r1r2r3r3≥27 (R)  a+b>c⇒2b−c+b>c⇒3b>2c⇒bc>23  since a+c=2bb+c>a⇒b+c>2b−c⇒bc<2 Again, c+a>b⇒2b>b∴bc∈23,2 (S) We have, PDAD=ar⁡(ΔBPC)ar⁡(ΔBAC),….etc. ∴PDAD+PEBE+PFCF=ar⁡(ΔBPC)+ar⁡(ΔCPA)+ar⁡(ΔAPB)ar⁡(ΔABC)=1