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.
Anonymous:
ohh yeah sure
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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:
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