CBSE Class 10 Answered
triangles
Asked by debasish2345 | 15 Jul, 2009, 05:42: PM
Expert Answer
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
so
AE/ED=AX/XF...(BPT)......Eqn (i)
simly,
in trianlge BCX,
BD/DC=XF/FC... eqn (ii)
the value of LHS of both (i) and (ii) is equal to 1(from given)
so
AX/XF=XF/FC=1
thus,
AX=XF=FC...(iii)
now consider triangle CDA,
GX II AD
so using BPT,
CX/XA=CG/GD
From (iii) we ge that,
CX=2AX
i.e.
CX/AX=2
So,
CG/GD=2
but, CG+GD=CD
So
CG=2/3[CD]=2/3[BC/2]=BC/3
DG=1/3[DC]=1/3[BC/2]=BC/6
BD=BC/2
so
BD/DG=3/1
In triangle GBX,
BE/EX=BD/DG (BPT)
so
BE/EX=BD/DG=3/1
hence proved.
Answered by | 28 Jul, 2009, 03:24: PM
CBSE 10 - Maths
Asked by s.sreeram | 11 Mar, 2010, 08:14: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by BhavSimran | 10 Mar, 2010, 12:24: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by BhavSimran | 10 Mar, 2010, 12:08: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by BhavSimran | 10 Mar, 2010, 12:05: PM
ANSWERED BY EXPERT