Asked by debasish2345 | 15th Jul, 2009, 05:42: PM
draw the figure as mentioned in the question.
now, thru' X, draw line parallel to AD meeting BC at G.
Thru' D draw a line parallel to BX meeting AC in F.
in triangle ADF,
EX parallel to DF
in trianlge BCX,
BD/DC=XF/FC... eqn (ii)
the value of LHS of both (i) and (ii) is equal to 1(from given)
now consider triangle CDA,
GX II AD
so using BPT,
From (iii) we ge that,
In triangle GBX,
Answered by | 28th Jul, 2009, 03:24: PM
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