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Quadrilaterals - An Important Property of Triangles: The Midpoint Theorem

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Midpoint Theorem states that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is equal to half of it.

Triangle ABC with points E and F as midpoints of AB and AC respectively, showing line segment EF.
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The Converse of the Midpoint Theorem: The line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side.

A line starting from the midpoint of one side of a triangle and drawn parallel to the base.
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In △ABC\triangle ABC, if DD and EE are midpoints of ABAB and ACAC, then DE∥BCDE \parallel BC and DE=12BCDE = \frac{1}{2} BC.

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A quadrilateral formed by joining the midpoints of the sides of any quadrilateral is always a parallelogram.

📐Formulae

DE∥BCDE \parallel BC

DE=12BCDE = \frac{1}{2} BC

AD=DB=12ABAD = DB = \frac{1}{2} AB

AE=EC=12ACAE = EC = \frac{1}{2} AC

💡Examples

Problem 1:

In △ABC\triangle ABC, DD, EE, and FF are the midpoints of sides ABAB, BCBC, and CACA respectively. If BC=8BC = 8 cm, AC=10AC = 10 cm, and AB=12AB = 12 cm, find the perimeter of △DEF\triangle DEF.

Triangle ABC with midpoints D, E, F joined to form a smaller triangle DEF.

Solution:

By the Midpoint Theorem, the segment joining the midpoints of two sides is half the third side. DE=12AC=12×10=5DE = \frac{1}{2} AC = \frac{1}{2} \times 10 = 5 cm EF=12AB=12×12=6EF = \frac{1}{2} AB = \frac{1}{2} \times 12 = 6 cm DF=12BC=12×8=4DF = \frac{1}{2} BC = \frac{1}{2} \times 8 = 4 cm Perimeter of △DEF=DE+EF+DF=5+6+4=15\triangle DEF = DE + EF + DF = 5 + 6 + 4 = 15 cm.

Explanation:

We apply the Midpoint Theorem three times to find the lengths of the sides of the inner triangle formed by the midpoints.

Problem 2:

In △PQR\triangle PQR, SS is the midpoint of PQPQ. A line through SS is drawn parallel to QRQR to intersect PRPR at TT. If PR=14PR = 14 cm, find PTPT.

Triangle PQR with line ST parallel to QR, where S is midpoint of PQ.

Solution:

In △PQR\triangle PQR, SS is the midpoint of PQPQ and ST∥QRST \parallel QR. According to the Converse of Midpoint Theorem, a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side. Therefore, TT is the midpoint of PRPR. PT=12PR=12×14=7PT = \frac{1}{2} PR = \frac{1}{2} \times 14 = 7 cm.

Explanation:

This problem uses the Converse of the Midpoint Theorem to prove that TT is a midpoint, thus calculating the segment length as half of the total side.