The Mid-point Theorem & its Converse

The Midpoint Theorem and its converse are fundamental concepts in geometry, especially when dealing with triangles. They help us relate the midpoints of the sides of a triangle to the third side and the lines connecting these midpoints.

Formulae

Midpoint Theorem: In a triangle, the line segment joining the midpoints of any two sides is parallel to the third side and is half its length.

Converse of the Midpoint Theorem: In a triangle, the line drawn from the midpoint of one side, parallel to another side, bisects the third side.

Examples

Example-1:

Consider triangle $ABC$. Let $D$ be the midpoint of side $AB$, and $E$ be the midpoint of side $AC$. According to the Midpoint Theorem, $DE \parallel BC$ and $DE = \frac{1}{2}BC$. If $BC = 10$ cm, then $DE = 5$ cm.

Example-2:

Consider triangle $PQR$. Let $M$ be the midpoint of $PQ$, and $MS$ be a line drawn parallel to $QR$. According to the converse of the Midpoint Theorem, $S$ must be the midpoint of $PR$. If $PS = 7$ units, then $SR = 7$ units.

Theorem with Proof

Midpoint Theorem: The line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.

Proof:

Let $ABC$ be a triangle, where $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$. We need to prove that $DE \parallel BC$ and $DE = \frac{1}{2}BC$.

1. Construction: Draw a line through $C$ parallel to $AB$, intersecting the extension of $DE$ at $F$.

2. Proof:

• In $\triangle ADE$ and $\triangle CFE$:
  • $\angle ADE = \angle ECF$ (Alternate interior angles, since $DE \parallel CF$)
  • $AE = CE$ (Since $E$ is the midpoint of $AC$)
  • $\angle DEA = \angle FEC$ (Vertically opposite angles)
• Therefore, $\triangle ADE \cong \triangle CFE$ by the Angle-Side-Angle (ASA) congruence criterion.
• Thus, $AD = CF$ and $DE = EF$ (Corresponding parts of congruent triangles are equal).
• But $AD = BD$ (Since $D$ is the midpoint of $AB$).
• So, $BD = CF$.
• Also, $BD \parallel CF$ (By construction, since $AB \parallel CF$).
• Hence, $BCFD$ is a parallelogram (because one pair of opposite sides, $BD$ and $CF$, are equal and parallel).
• In a parallelogram, opposite sides are parallel. So, $DE \parallel BC$.
• Finally, since $DE = EF$ and $DE + EF = DF$, therefore, $DE = \frac{1}{2}DF$. Since $DF = BC$ in a parallelogram, we conclude $DE = \frac{1}{2}BC$.

Common mistakes by students

• Incorrectly Applying the Theorem: Students often struggle to correctly identify which line segment is bisected or which is parallel to which, leading to incorrect calculations.
• Forgetting the Half-Length Relationship: Students may remember that the line segment joining the midpoints is parallel but forget to include the “half the length” part ($DE = \frac{1}{2}BC$).
• Assuming Midpoints Without Proof: They may assume a point is a midpoint without the necessary information to prove it, leading to flawed logic.
• Confusing Converse with Theorem: Students might mix up the conditions of the Midpoint Theorem with the converse.

Real Life Application

The Midpoint Theorem and its converse have practical applications in various fields:

• Engineering: Used in designing structures, determining the stability of triangular frameworks.
• Surveying and Land Measurement: Helps to quickly determine distances and layout designs in land surveying.
• Architecture: Used for calculating the dimensions of architectural elements, such as bracing in triangular roof supports.

Fun Fact

The Midpoint Theorem can be generalized to higher dimensions. For instance, in a tetrahedron (a 3D triangle), the line joining the midpoints of any two edges is parallel to the opposite edge, and its length is half the length of that edge.

Recommended YouTube Videos for Deeper Understanding

Q.1 In triangle $ABC$, $D$ and $E$ are the midpoints of sides $AB$ and $AC$ respectively. If $DE = 5$ cm, then $BC$ is:

Check Solution

Ans: C

By the midpoint theorem, $DE = \frac{1}{2} BC$. Therefore, $BC = 2 \times DE = 2 \times 5 = 10$ cm.

Q.2 In triangle $PQR$, $X$ is the midpoint of $PQ$ and $Y$ is a point on $PR$ such that $XY$ is parallel to $QR$. If $PR = 12$ cm, then $PY$ is:

Check Solution

Ans: B

Since $X$ is the midpoint of $PQ$ and $XY \parallel QR$, by the converse of the midpoint theorem, $Y$ is the midpoint of $PR$. Thus, $PY = \frac{1}{2} PR = \frac{1}{2} \times 12 = 6$ cm.

Q.3 In triangle $LMN$, $K$ and $J$ are midpoints of $LM$ and $LN$ respectively. If the area of triangle $LKJ$ is 8 $cm^2$, what is the area of triangle $LMN$?

Check Solution

Ans: C

By the midpoint theorem, $KJ \parallel MN$ and $KJ = \frac{1}{2} MN$. The triangles $LKJ$ and $LMN$ are similar, and the ratio of their corresponding sides is 1:2. The ratio of their areas is the square of the ratio of their sides, which is $1^2 : 2^2$ or 1:4. If area of $LKJ$ is 8, area of $LMN$ will be $8 \times 4 = 32$ $cm^2$.

Q.4 In triangle $ABC$, $D$ and $E$ are points on sides $AB$ and $AC$ respectively. If $AD = 3$ cm, $DB = 4$ cm, $AE = 4.5$ cm, and $EC = 6$ cm, is $DE \parallel BC$?

Check Solution

Ans: A

We use the converse of the Basic Proportionality Theorem, which is closely related to the midpoint theorem. If $AD/DB = AE/EC$, then $DE \parallel BC$. Here, $AD/DB = 3/4$ and $AE/EC = 4.5/6 = 3/4$.

Q.5 The sides of a triangle are 6 cm, 8 cm and 10 cm. If $D$, $E$, and $F$ are the midpoints of the sides, what is the perimeter of the triangle formed by joining $D$, $E$, and $F$?

Check Solution

Ans: B

The triangle formed by joining the midpoints has sides equal to half the length of the original triangle’s sides. So the new triangle has sides 3 cm, 4 cm, and 5 cm. Perimeter = 3 + 4 + 5 = 12 cm.

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