CBSE Class 10 Answered

we see that

 in triangles

ECF and AFG,

angle CEF=angle FAG..each is right angle

angle ECF=AFG...corr angles with FG II CB

so the triangles ECF,AFG are similar.....(i)

simly we can show thgat

 triangles

DGB,AFG are similar...(II)

so from (i) and (ii)

ECF,DGB are similar triangles.

so

EC/CF=DG/GB=EF/DB

i.e.

EC/DG==CF/GB=EF/DB

i.e.

EC X BD=EF X DG...(iii)

but

DEFG is a square(given)

so

DE=EF=FG=DG

so from (iii) we get

EC X BD= DE X DE

hence proved.

DG/GB=...(ii)

from (i) and (ii)

EC/CF=DB/AG