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Question
In ΔABC, D and E are the midpoints of the sides AB and AC respectively. F is any point on the side BC. If DE intersects AF at P show that DP = PE.
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Solution
Note: This question is incomplete.
According to the information given in the question,
F could be any point on BC as shown below:
So, this makes it impossible to prove that DP = DE, since P too would shift as F shift because P too would be any point on DE as F is.
Note: If we are given F to be the mid-point of BC, the result can be proved.
D and E are the mid-points of AB and AC respectively.
DE || BC and DE = `(1)/(2)"BC"`
But F is the mid-point of BC.
⇒ BF = FC = `(1)/(2)"BC"` = DE
Since D is the mid-point of AB, and DP || EF, P is the mid-point of AF.
Since P is the mid-point of AF and E is the mid-point of AC,
PE = `(1)/(2)"FC"`
Also, D and P are the mid-points of AB and AF respectively.
⇒ DP = `(1)/(2)"BF"`
= `(1)/(2)"FC"`
= PE ....(Since BF = FC)
⇒ DP = PE.
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