this theorem is called as fundamental theorem or basic proportionality theorem or BPT theorem or theorem suggested by thalas so it can be asked by any of the name
statement : A LINE IS DRAWN PARALLEL TO ONE SIDE OF TRIANGLE TO INTERSECT OTHER TWO SIDES IN TWO DISTINCT POINTS, THE OTHER TWO SIDES ARE DIVIDED IN SAME RATIO
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Verified answer
Answer:
Given- D and E are the midpoints of side AB and AC respectively
DE is parallel to BC
To prove- AD/AB = AE/AC
Proof- AD = DB ----- (By midpoint theorem)
Similarly AE = EC ----- (By midpoint theorem)
Therefore,
2(AD) = AB ----- (as D is the midpoint) ---- eq.1
Similarly, 2(AE) = AC ----- (as E is the midpoint) ---- eq.2
Substituting eq.1 and eq.2, we get
AD/AB = AE/AC
here's your answer
this theorem is called as fundamental theorem or basic proportionality theorem or BPT theorem or theorem suggested by thalas so it can be asked by any of the name
statement : A LINE IS DRAWN PARALLEL TO ONE SIDE OF TRIANGLE TO INTERSECT OTHER TWO SIDES IN TWO DISTINCT POINTS, THE OTHER TWO SIDES ARE DIVIDED IN SAME RATIO
for other detail refer to attachment
thank you