State the mid-point theorem and its converse....URGENT
Answers
Proof
The proof of mid point theorem is as follows.
Have a look at the following diagram:

Here, In $\bigtriangleup$ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined.
Given: AD = DB and AE = EC.
To Prove: DE $\parallel$ BC and DE = $\frac{1}{2}$ BC.
Construction: Extend line segment DE to F such that DE = EF.
Proof: In $\bigtriangleup$ ADE and $\bigtriangleup$ CFE
AE = EC (given)
$\angle$AED = $\angle$CEF (vertically opposite angles)
DE = EF (construction)
hence
$\bigtriangleup$ ADE $\cong$ $\bigtriangleup$ CFE (by SAS)
Therefore,
$\angle$ADE = $\angle$CFE (by c.p.c.t.)
$\angle$DAE = $\angle$FCE (by c.p.c.t.)
and AD = CF (by c.p.c.t.)
The angles $\angle$ADE and $\angle$CFE are alternate interior angles assuming AB and CF are two lines intersected by transversal DF.
Similarly, $\angle$DAE and $\angle$FCE are alternate interior angles assuming AB and CF are two lines intersected by transversal AC.
Therefore, AB $\parallel$ CF
So, BD $\parallel$ CF
and BD = CF (since AD = BD and it is proved above that AD = CF)
Thus, BDFC is a parallelogram.
By the properties of parallelogram, we have
DF $\parallel$ BC
and DF = BC
DE $\parallel$ BC
and DE = $\frac{1}{2}$BC (DE = EF by construction)
Hence proved.
Converse
The converse of midpoint theorem does also exist. In geometry, the converse of midpoint theorem is as important as the theorem itself.
The converse of midpoint theorem states that:
"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.
Explanation:
Have a look at the figure given below:
Consider the following triangle $\bigtriangleup$ABC, in which D is the midpoint of AB. A line segment DE is drawn which meets AC in E and is
parallel to the opposite side BC. In this case, the third AC is bisected by the line segment DE; i.e. we can say that
if AD = DB and DE $\parallel$ BC
then AE = EC.
MidPoint Theorem Statement
The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”
Construction- Extend the line segment DE and produce it to F such that, EF=DE.
In the triangle, ADE, and also the triangle CFE
EC= AE —– (given)
∠CEF = ∠AED {vertically opposite angles}
EF = DE { by construction}
hence,
△ CFE ≅ △ ADE {by SAS}
Therefore,
∠CFE = ∠ADE {by c.p.c.t.}
∠FCE= ∠DAE {by c.p.c.t.}
and CF = AD {by c.p.c.t.}
The angles, ∠CFE and ∠ADE are the alternate interior angles. Assume CF and AB as two lines which are intersected by the transversal DF.
In a similar way, ∠FCE and ∠DAE are the alternate interior angles. Assume CF and AB are the two lines which are intersected by the transversal AC.
Therefore, CF ∥ AB
So, CF ∥ BD
and CF = BD {since BD = AD, it is proved that CF = AD}
Thus, BDFC forms a parallelogram.
By the use of properties of a parallelogram, we can write
BC ∥ DF
and BC = DF
BC ∥ DE
and DE = (1/2 * BC).
Hence, the midpoint theorem is Proved.
and
The converse of MidPoint Theorem
The converse of the midpoint theorem states that ” if a line is drawn through the midpoint of one side of a triangle, and parallel to the other side, it bisects the third side”.Given: In ∆PQR, S is the midpoint of PQ, and ST is drawn parallel to QR.
Converse of Midpoint Theorem Proof
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To prove: ST bisects PR, i.e., PT = TR.
Construction: Join SU where U is the midpoint of PR.
Converse of Midpoint Theorem
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Proof:
Statement
Reason
1. SU ∥ QR and SU = 12QR.
1. By Midpoint Theorem.
2. ST ∥QR and SU ∥ QR.
2. Given and statement 1.
3. ST ∥ SU.
3. Two lines parallel to the same line are parallel themselves.
4. ST and SU are not the same line.
4. From statement 3.
5. T and U are coincident points.
5. From statement 4.
6. T is the midpoint of PR (Proved).
6. From statement 5.
