Question

converse of mid point theorem with full solution.......​

Answer 1

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$\huge\mathfrak\pink {Question :}$

Converse of MID-POINT theorem with full solution.

$\huge\mathfrak\purple {Answer :}$

$\large\bold\red {Converse\:of\:MID-POINT\: theorem :}$

$\huge\mathfrak\blue {Solution :}$

The line drawn through the mid-point of one side of a triangle, parallel to another side, intersects the third side at its mid-point.

$\large\green {Given:}$

• In triangle ABC, D is the mid-point of AB.

• DE || BC and DE intersects AC at point E

$\large\green {To\: prove:}$

E is the mid-point of AC

$\large\green {Proof:}$

• Let E not be the mid-point of AC, and let F be the mid-pointof AC.

• Now, in triangle ABC, D is the mid-point of AB and F is the mid-point of AC.

• Therefore, DF || BC ..... (i)

• But DE || BC ..... (ii)

• From (i) and (ii) it is clear that two intersecting lines DE and DF are parallel to BC. But this is in contradiction to parallel line axiom.

• Therefore our supposition that E is not the mid-point of AC is wrong.

• Hence E is the mid-point of AC.

$\large\bold {This\:was\:to\:be\:proved.}$

(Figure is in the attachment.)

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Answer 2

$\huge\bf\mathscr\pink{Your\: Answer}$

step-by-step explanation:

Converse of midpoint theorem :

"If a line segment is drawn passing through the midpoint of any one side of a triangle and parallel to another side, then this line segment bisects the remaining third side.

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 || 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' || BC.

But we have DE || 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.

Hence,

proved✍️✍️