Question

In a triangle ABC (with AC as base,and the name of the vertices are given in anticlockwise manner) ,BC is produced to D , from D a line is drawn intersecting AC and AB at E and F respectively. E is the midpoint of AC and angleAEF=angleAFE.
prove that BD/CD=BF/CE.

Answer 1

Answer:

AE=CE.........1

AE=AF(Sides opp to equal angles)....(2)

ang AFE=ang AEF.....(3)

from (1) and (2)

AE=CE=AF......(4)

draw CG parallel to DF to intersect AB at G

ang AFE=ang AGC(corresponding angles)....(5)

ang AEF=ang ACG......(6)

from 3,5,6

ang ACG=ang AGC

AC=AG.....(7)

AE+EC=AF+FG

from 4,7 we get

AE=EC=AF=FG

GF=EC=AF....(8)

by thales theorem

BC/CD=BG/GF

adding 1 on both sides we get

BC+CD/CD=BG+GF/GF

BD/CD=BF/GF

from 8 we get BD/CD=BF/CE

Step-by-step explanation: