Question

state an prove bpt theorem in detail....?????

Answer 1

Given - In triangle ABC,  DE is parallel to BC. DE bisects AB at D and AC at E.

To Prove - AD/BD= AE/CE

Construction - Join BE and CD and draw EF is perpendicular to AB and DG is perpendicular to AC.

Proof - 

ar(ADE)=1/2×B ×H
1/2×AD×EN---------(1)
ar(BDC)=1/2×B ×H
1/2× DB ×EN---------(2)
dividing 1 and 2
ar(ADE)/ar(BDE)=1/2×AD×EN/1/2×DB×EN-----(1)
ar(ADE)=1/2×B ×H
1/2×AE×DE----(3)
ar(DEC)=1/2×B×H
1/2×EC×DM--------(4)
divide 3&4
arADE/arBEC=1/2×AE × DM/1/2×EC×DM
arADE/arDEC=AE/EC
now,BDE and DEC r on the same base between same parellel line bc&de
ar BDE =ar DEC
hence
ar ADE/ar BDE = ar ADE / ar DEC
AD/DB = AE / EC
hope its help u

Answer 2

hope it will help uh ...
nd plz mark as brainliest ...