Question

please answer my problem no. 6​

Answer 1

Answer:

hii mate. :-

your solution:-

Step-by-step explanation:

Answered

'd' is the midpoint of side 'bc' of triangle 'abc'.'ad' is bisected at point 'e' and 'be 'produced cuts 'ac' at point '

Expert Answer:

draw II from E to AC meeting BC in Z (say).

so , in triangle BCX

BE:EX=BZ:ZC.....BPT...(i)

now, in trianlgle ADC,

EZ II AC,..mentioned above

So,

DE:EA=DZ:ZC

But DE:EA=1:1...as E is mid pt of AD is given

so

DZ:ZC=1:1...(ii)

also,

BD:DC=1:1...as D is mid pt of BC given

Thus,DZ=DC/2=BD/2

So,

BZ:ZC=(BD+DZ):ZC=(2DZ+DZ):ZC=3DZ:ZC=3ZC:ZC=3:1...(iii)...using (ii)

So from (i) and (iii)

we get,

BE:EX=BZ:ZC=3:1

Hence proved...