36I hope this helps
I m just doing it using Thales' theorem... the only one i know for these sort of sums...
1>
we draw the figure as per the question. Now,
AF/BF = AE/CE = 1/3
so, EF || BC [by Thales' theorem]
=> ∆EOF ~ ∆BOC
=> OB/OE = BC/EF
Now, clearly, ∆AEF ~ ∆ACB [since EF || BC]
so, AB/AF = BC/EF
now, BF/AF = 3 => BF/AF + 1 = 3 + 1 => (BF + AF)/AF = 4 => AB/AF = 4
=> BC/EF = 4
Thus, OB/OE = 4 ...
Next comes,
OD/OA [the biggest tension of this problem]
Again, do this construction inside the triangle and name the points...
clearly, ∆OEQ ~ ∆OBD [by AAA]
so, OD/OQ = OB/OE = 4
=>(OD + OQ)/OQ = 5
=> DQ/OQ = 5 ---- (i)
and, AQ/QD = 1/3 ---- (ii)
(i) x (ii)
AQ/OQ = 5/3
=> (AQ + OQ)/OQ = (5 + 3)/3
=> AO/OQ = 8/3 ---- (iii)
and, OQ/OD = 1/4 --- (iv)
on, (iii) x (iv) we have,
OA/OD = 2/3 Ans..
Hence, OB/OE = 4 and OD/OA = 3/2
sorry.. if there r lill errors... and i know it must me wrong... coz i've solved it...
