BE and CF are altitudes of triangle ABC. D is the midpoint of BC. Prove that DE=DF.
Answers
Given : BE and CF are altitudes of triangle ABC. D is the midpoint of BC.
To find : Prove that DE=DF.
Solution:
BE and CF are altitudes of triangle ABC
=> ΔBCE & ΔBCF are right angle triangles on the same same
Two right angle triangles on the same base
means that Base is the diameter of the circle
on which points of right triangles lies
Hence BC is diameter
D is the mid point of BC
Hence D is the center of circle
Hence DE & DF are the radius of circle
Hence DE = DF = Radius
QED
Hence Proved
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BE and CF are altitudes of triangle ABC
=> ΔBCE & ΔBCF are right angle triangles on the same same
Two right angle triangles on the same base
means that Base is the diameter of the circle
on which points of right triangles lies
Hence BC is diameter
D is the mid point of BC
Hence D is the center of circle
Hence DE & DF are the radius of circle
Hence DE = DF = Radius
QED
Hence Proved
Step-by-step explanation: