triangles

Asked by debasish2345 | 15th 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  | 28th Jul, 2009, 03:24: PM

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