# The Mid Point Theorem

**Theorem 6Â **

In a triangle, the line segment joining the mid points of any two sides is parallel to the third side.

Let us look at the different types of quadrilateralsÂ drawn below.

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**Given:** In Î”ABC, D is the mid-point of AB and E is the mid-point of AC.

**To prove:** DE = BC and DE Ã§Ã§ BC

**Construction:** Produce line-segment DE to F such thatÂ DE = EF and join FC.

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**Proof:**

In Î”ADE and Î”CFEÂ

AE = ECÂ Â Â Â Â Â ...(given)

DE = EFÂ Â Â Â Â ...(by construction)

âˆ AED = âˆ CEFÂ Â Â Â Â Â Â ...(Vertically opposite âˆ s)

.^{.}. Î”ADE â‰… Î”CFE

.^{.}. AD = CFÂ Â Â Â ...(i)

and âˆ ADE = âˆ CFEÂ ...(CPCT)

SinceÂ AD = DBÂ ...(D is the mid-point)

andÂ Â Â AD = CFÂ Â Â ...(proved above)

.^{.}. DB = FCÂ Â Â Â Â Â Â ...(ii)

Since âˆ ADE = âˆ CFE

.^{.}.Â AD Ã¯Ã¯ CF .^{.}.Â Â AB Ã¯Ã¯ CFÂ Ãž DB Ã¯Ã¯ CFÂ ...(iii)

Now in quadrilateral BCFD,

DB = FC and DB Ã¯Ã¯ FC

.^{.}. BCFD is a parallelogram

BC = DF = DE + EF = 2 DE

.^{.}.Â DE = BC

Also,BC Ã¯Ã¯ DFÂ ...(opposite sides of ||gm)

.^{.}. BC Ã¯Ã¯ DEÂ ÃžÂ Â DE Ã¯Ã¯ BC

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In the figure M, N and O are the mid-points of sides PR, PQ and RQ of Î”PQR and PQ = QR = PR. Show that Î”MON is an equilateral triangle.

Since M and N are the mid-points of PR and PQ of Î”PQR.

.^{.}.Â MN = RQ and MN Ã¯Ã¯ RQ...(i)

Similarly, we can get

MO = PQ and MO Ã¯Ã¯ PQ...(ii)

NO = PR and NO Ã¯Ã¯ PRÂ Â ...(iii)

ButÂ PQ = QR = PR...(given)

.^{.}.Â PQ = QR = PR

.^{.}.Â MO = MN = NO

Hence, Î”MNO is an equilateral triangle.Â Â