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May 15, 2024**Midpoint Theorem and Converse of Midpoint Theorem in Triangle:** Geometry is the branch of mathematics that deals with various shapes and objects. Of all, Triangles are one of the most important topic in geometry. A triangle is the smallest polygon made up of three line segments: midpoint theorem and converse of midpoint theorem deal with the midpoints of the triangle. A midpoint is the middle point of a line segment which is equidistant from both its ends. Midpoint theorem is used in the field of coordinate geometry, calculus, and algebra also.

Let us discuss the statements, proofs of the midpoint theorem, and converse of the midpoint theorem with the help of examples in this article.

**Learn the Concepts of Midpoint Theorem**

The midpoint theorem states that a line connecting the midpoints of any two sides of a triangle is parallel to the other side of the triangle and has a value equal to half of the value of the other parallel side.

The midpoint theorem is one of the applications of the basic proportionality theorem or Thales theorem, which is also used in coordinate geometry, calculus, and algebra.

The converse of the midpoint theorem is the statement of the triangle’s midpoint theorem. The converse of the midpoint theorem states that a line drawn parallel to a side from one side’s midpoint to the other side bisects the other side.

The application of the converse of the basic proportionality theorem is the converse of the midpoint theorem.

Midpoint theorem tells the relation between the line joins between the midpoints of any two sides of the triangle to the other side of the triangle.

**Statement: **In a triangle, if a line joining the midpoints of any two sides of the triangle is parallel to the third side of the triangle, and its value equals half of the measure of the third side of the triangle.

**To Prove:** We have to prove \(DE\) is parallel to \(BC\) and equals half of the \(BC.\)

**Proof:** Let us study the proof for describing the midpoint theorem in triangles. Consider a triangle (Say \(△ABC\)),

Let the points \(D\) and \(E\) be the midpoints of the side \(AB\) and \(AC\) of triangle \(△ABC.\)

\(AD=DB\) and \(AE = EC \cdots \cdots {\rm{(i)}}\)

\(DE\) is the line drawn from the midpoints of the sides \(AB\) and \(AC\) of the triangle \(△ABC.\)

We have to prove the line joining the midpoints \(D,\) and \(E\) of the sides \(AB\) and \(AC\) of the triangle \(ABC\) is parallel to the third side \(BC.\)

Extend the line* *\(DE\) till point \(F,\) such that \(DE\) equals \(EF.\)

Join the points \(C\) and \(F.\)

Here, the lines \(AB\) and \(CF\) are parallel, and the line \(AC\) is transversal.

We know that the pair of alternate interior angles formed by two parallel lines by a transversal are equal.

So, the angles \(DAE\) and \(ECF\) are equal, as they are alternate interior angles formed by the parallel lines \(AB\) and \(CF\) with transversal \(AC.\)

\(\angle DAE = \angle ECF\) (Alternate interior angles)

Now, in triangles \(ADE\)* *and \(ECF,\)

Sides \(AE=EC\) (Given \(E\) is the midpoint of side \(AC\))

\(\angle AED = \angle CEF\) (Vertically opposite angles)

\(\angle DAE = \angle ECF\) (Alternate interior angles)

By the A.S.A congruency rule, two triangles \(ADE\) and \(CEF,\) are congruent.

\(△ADE≅△CEF\)

\(AF=DE\) (By C.P.C.T)

And \(CF=AD\) (By C.P.C.T)

From \({\rm{(i),}}\) \(BD=CF\)

Also, by construction, \(BD\) is parallel to \(CE.\)

Hence, \(BDFC\) is a parallelogram.

That means \(DF∥BC,\) which implies \(DE\) is parallel to \(BC.\)

We know that \(DE = EF = \frac{1}{2}DF = \frac{1}{2}BC\)

Therefore, we can say the line joining the midpoints (\(D\) and \(E\)) of the sides \(AB\) and \(AC\) is parallel to the other side \(BC\) and equals half of the \(BC.\)

\(DE∥BC\)* *and \(DE = \frac{1}{2}B\)

This theorem is the converse proof of the mid-point theorem of triangles.

**Statement: **The line drawn through the midpoint of any side of the triangle parallel to the other side of the triangle bisects the third side of the triangle.

**To Prove:** We have to prove the point \(E\) is the midpoint of the side \(AC (AE=EC).\)

**Proof:** Let \(D\) be the midpoint of the side \(AB\) of triangle \(△ABC.\)

\(AD = BD \cdots \cdots {\rm{(i)}}\)

The line \(DE\) is drawn from point \(D\) to the side \(AC,\) which is parallel to the side \(BC.\)

From point \(C,\) draw a line \(CM\) parallel to \(AB.\)

Extends the line \(DE,\) which intersects the line \(CM\) at \(F.\)

Given \(DE∥BC,\) that implies \(DF∥BC\) and \(BD∥CF\) (By construction)

Thus, the quadrilateral \(BDFC\) is a parallelogram.

\(BD=CF\) (By C.P.C.T)

From i, \(AD=CF\)

By construction, the lines \(BD\) and \(CF\) are parallel, and the line \(DF\) is the transversal.

The pair of alternate angles formed by the parallel lines with the transversal are equals.

Thus, \(\angle ADE = \angle CFE\)

Now, in triangles \(ADE\)* *and* *\(ECF,\)

\(AD=CF\) (Proved above)

\(\angle ADE = \angle CFE\) (Alternate interior angles)

\(\angle AED = \angle CEF\) (Vertically opposite angles)

By the A.S.A congruency rule, two triangles \(ADE\) and \(CEF,\) are congruent.

\(△ADE≅△CEF\)

The sides \(AE=EC\) (By C.P.C.T)

Therefore, point \(E\) is the midpoint of the side \(AC.\)

** Q.1. **\(l, m,\)

** Ans:** Given \(AB:BC=1:1\)

Join \(AF,\) which intersects the line m at point \(G.\)

In \(△ACF,\)

\(AB:BC=1:1,\) which means \(B\) is the midpoint of \(AC.\)

And, \(BG∥CF,\) as given, the lines \(m\) and \(n\) are parallel.

By the converse of midpoint theorem, the line drawn from the midpoint parallel to the other side bisects the third side of the triangle.

Therefore, \(G\) is the midpoint of side \(AF.\)

Thus, \(AG=GF.\)

Now, in \(△AFD,\)

\(AG=GF\) (Proved above) and \(GE∥AD\) as the lines l and m are parallel.

Therefore, by the converse of midpoint theorem, \(E\) is the midpoint of side \(DF.\)

Thus, \(DE=EF\)

Hence, the ratio \(DE: EF=1:1.\)

** Q.2. Find the perimeter of the triangle formed by joining **\(L, M,\)

** Ans:** Given that \(L, M\) and \(N\) are the midpoints of the sides \(PQ, QR,\) and \(PR\) of triangle \(PQR.\)

Given, \(PQ = 8\;{\rm{cm}},QR = 9\;{\rm{cm}}\) and \(PR = 6\,{\rm{cm}}{\rm{.}}\)

By midpoint theorem, the line joining the midpoints of any two sides is parallel to the third side and equals half of the third side.

In triangle \(PQR,\)

\(LM∥PQ\) and \(LM = \frac{1}{2}PQ = \frac{1}{2}(8) = 4\;{\rm{cm}}\)

\(LN∥PR\) and \(LN = \frac{1}{2}PR = \frac{1}{2}(6) = 3\;{\rm{cm}}\)

\(MN∥QR\) and \(MN = \frac{1}{2}QR = \frac{1}{2}(9) = 4.5\;{\rm{cm}}\)

** Q.3. The values given in the below figure are indicated in centimetres. Length of the side **\(AX = 9\,{\rm{cm}}{\rm{.}}\)

** Ans:** From the given figure,

\(AD = DB = 4\;{\rm{cm}}\) and \(AE = EC = 6\;{\rm{cm}}\)

Therefore, points \(D\) and \(E\) are the midpoints of the sides \(AB\) and \(AC.\)

By midpoint theorem, the line joining the points \(D\) and \(E\) is parallel to side \(BC\) and equals half of \(BC.\)

\(DE∥BC,\) which implies \(DO∥BX.\)

Now, in triangle \(ABX,\) the line \(DO\) is drawn from the midpoint \(D\) of the side \(AB\) and parallel to the side \(BX.\)

By converse of midpoint theorem, we know that the line is drawn from the midpoint of one side parallel to the other side and bisects the third side of the triangle.

Thus the line \(DO\) bisects the third line \(AX.\)

\(AO = \frac{1}{2}AX = \frac{1}{2}(9\;{\rm{cm}}) = 4.5\;{\rm{cm}}\)

Hence, the value of \(AO\) is \(4.5\;{\rm{cm}}.\)

** Q.4. In a triangle, **\(ABC, X\)

** Ans:** Given, \(AB = 10,AC = 8\) and \({\rm{BC = 6\, units}}{\rm{.}}\)

Also, given that points, \(X\) and \(Y\) are the midpoints of sides \(AC\) and \(BC.\)

By midpoint theorem, the line joining the midpoints of any two sides is parallel to the third side and equals half of it.

So, the line \(XY\) is parallel to \(AB\) and equals half of \(AB.\)

\(XY = \frac{1}{2}AB = \frac{1}{2}(10) = 5\,{\rm{units}}\)

** Q.5. Find the value of the hypotenuse of the triangle **\(ABC,\)

** Ans:** Given, \(AB = 10,AC = 8\) and \({\rm{BC = 6 \,units}}{\rm{.}}\)

Length of the line segment joining the midpoints \(X\) and \(Y\) of the sides \(AC\) and \(BC\) are \(5\) units.

By midpoint theorem, the line joining the midpoints of any two sides is parallel to the third side and equals half of it.

So, the line \(XY\) is parallel to \(AB\) and equals half of \(AB.\)

\(XY = \frac{1}{2}AB\) units

\({\rm{AB = 2 XY = 2(5) = 10 \.units}}\)

In this article, we have studied the statement and proof of midpoint theorem and the statement and proof of the converse of midpoint theorem. The midpoint theorem states that in a triangle, if a line joining midpoints of any two sides of the triangle is parallel to the third side of the triangle, and its value equals half of the measure of the third side of the triangle.

The converse of mispoint theorem states that the line drawn through the midpoint of any side of the triangle parallel to the other side of the triangle bisects the third side of the triangle. This article gives the solved examples on the midpoint theorem and converse of midpoint theorem, which help us to understand easily.

*Q.1. What is the midpoint?*** Ans:** The midpoint is the point on the line segment equidistant from both ends of the line segment.

*Q.2. Define midpoint theorem?*** Ans:** In a triangle, if a line joining midpoints of any two sides of the triangle is parallel to the third side of the triangle, and its value equals half of the measure of the third side of the triangle.

*Q.3. What is the converse of the midpoint theorem?*** Ans:** The line drawn through the midpoint of any side of the triangle parallel to the other side of the triangle bisects the third side of the triangle.

*Q.4. Is the midpoint theorem is an application of Thales theorem?*** Ans:** Yes. midpoint theorem is the application of Thales theorem.

*Q.5. Is the converse of midpoint theorem is the application of the converse of B.P.T.*** Ans:** Yes. The converse of midpoint theorem is the application of converse of B.P.T.