What is the Converse of Midpoint Theorem?
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Converse of a mid point theorem is the straight line drawn through the midpoint of one side of a triangle parallel to another bisects the third one.
If BCFD is a parallelogram
Cf=Bd( opposite sides are equal)
CF=Da as Bd=Da
In triangle ADE and CFE
Ad=Cf
angle DAE= angleEFC alternate angles
angleADE= angleECF alternate
triangle ADE congurent to triangle CFE by ASA
therefore Ae =Ec by cpctc
Thanq...
If BCFD is a parallelogram
Cf=Bd( opposite sides are equal)
CF=Da as Bd=Da
In triangle ADE and CFE
Ad=Cf
angle DAE= angleEFC alternate angles
angleADE= angleECF alternate
triangle ADE congurent to triangle CFE by ASA
therefore Ae =Ec by cpctc
Thanq...
quaz7:
Hi! Thanks for the answer. It was informative :)
Answered by
1
Converse Of Midpoint Theorem
Theorem: The line through the midpoint of one side of a triangle when drawn parallel to a second side bisects the third side.
Prerequisites:
Midpoint theorem (proof)
Unique parallel through a point (proof)
Proof:
Let ABC be a triangle such that D is the midpoint of the line AB. Let E be a point on the line AC, such that DE \parallel BC.
If possible, let us assume that E is not the midpoint of the line AC. Let E' be the midpoint of the line AC. Let us draw a line DE'.
Now in \triangle ABC, DE' is the line joining the midpoints of the lines AB and AC. Therefore, by Midpoint theorem, DE' \parallel BC.
But we have DE \parallel BC. This is a contradiction, as only a unique line can pass through the point D which is parallel to the line BC.
Hence, E is the midpoint of the line AC, i.e., line DE bisects the side AC.
Theorem: The line through the midpoint of one side of a triangle when drawn parallel to a second side bisects the third side.
Prerequisites:
Midpoint theorem (proof)
Unique parallel through a point (proof)
Proof:
Let ABC be a triangle such that D is the midpoint of the line AB. Let E be a point on the line AC, such that DE \parallel BC.
If possible, let us assume that E is not the midpoint of the line AC. Let E' be the midpoint of the line AC. Let us draw a line DE'.
Now in \triangle ABC, DE' is the line joining the midpoints of the lines AB and AC. Therefore, by Midpoint theorem, DE' \parallel BC.
But we have DE \parallel BC. This is a contradiction, as only a unique line can pass through the point D which is parallel to the line BC.
Hence, E is the midpoint of the line AC, i.e., line DE bisects the side AC.
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