Prove the converse of mid-point theorem.

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Solution

The converse of mid-point theorem: It states that in a triangle, line drawn from the mid-point of the one side of a triangle, parallel to the other side intersects the third side at its mid-point.
Given: $A B C$ is a triangle and $D$ is the mid-point of $A B .$
from $D$ a line $D E$ is drawn parallel to $B C,$ intersect $A C$ at $E .$
To prove: $E$ is the mid-point of $A C .$
Construction: Extend $D E .$ From $C$ draw a line $C F$ parallel to $B A,$ which intersect produced $D E$ at $F .$
Proof: Since $B D$ is parallel to $C F$ (by the construction)
and $D F$ is parallel to $B C$ (given)
So, $B D F C$ is a parallelogram.
$B D = C F$ [opposite sides of the parallelogram are equal]
$A D = B D$ [ $D$ is the mid-point of $A B$ ]
$A D = C F \dots \dots (i)$
In the triangle $A E D$ and $C E F,$ we have
$∠ A E D = ∠ C E F$ [Vertically opposite angles]
$∠ A D E = ∠ E F C$ [As $A B ∥ C F$ ]
$A D = C F$ [from $e q . (i)$ ]
$∴ △ A E D ≅ △ C E F$ [ by the AAS congruency rule]
$A E = E C$ [CPCT]
$i . e .$ $E$ is the mid-point of $A C .$

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Similar questions

Q. State and prove mid point theorem.

Mid point theorem is a special case of converse of basic proportionality theorem. (T/F)

Q. Prove the converse of the mid-point theorem following the guidelines given below: Consider a triangle

A B C

with

D

as the mid-point of

A B

. Draw

D E ∥ B C

to intersect

A C

E

. Let

E_{1}

be the mid-point of

A C

. Use mid-point theorem to get

D E_{1} ∥ B C

and

D E_{1} = B C / 2

. Conclude

E = E_{1}

and hence

E

is the mid-point of

A C

Question 8
Using converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

In the given figure, $D$ and $E$ are mid points of sides $A B$ and $A C$ of $△ A B C .$ Which of the following is correct?

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