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Geometry - Mid-point Theorem and its Converse

Grade 9ICSE

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

🔑Concepts

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The Mid-point Theorem states that the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. In △ABC\triangle ABC, if DD and EE are mid-points of ABAB and ACAC respectively, then DE∥BCDE \parallel BC and DE=12BCDE = \frac{1}{2} BC.

Triangle ABC with midpoints D and E joined by segment DE parallel to BC
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The Converse of the Mid-point Theorem states that the line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side. In △ABC\triangle ABC, if DD is the mid-point of ABAB and DE∥BCDE \parallel BC, then EE is the mid-point of ACAC (i.e., AE=ECAE = EC).

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The Medial Triangle is the triangle formed by joining the mid-points of the three sides of a triangle. This divides the original triangle into four congruent triangles. The area of the medial triangle is exactly 14\frac{1}{4} of the area of the original triangle.

Triangle with its medial triangle formed by connecting midpoints
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For any quadrilateral, the figure formed by joining the mid-points of its consecutive sides is always a parallelogram.

📐Formulae

If DD and EE are mid-points of ABAB and ACAC, then DE∥BCDE \parallel BC

Length of segment: DE=12BCDE = \frac{1}{2} BC

In △ABC\triangle ABC, if AD=DBAD = DB and DE∥BCDE \parallel BC, then AE=ECAE = EC

PerimeterofMedial△DEF=12(AB+BC+CA)Perimeter of Medial \triangle DEF = \frac{1}{2} (AB + BC + CA)

AreaofMedial△DEF=14×Area of △ABCArea of Medial \triangle DEF = \frac{1}{4} \times \text{Area of } \triangle ABC

💡Examples

Problem 1:

In △ABC\triangle ABC, the mid-points of sides AB,BCAB, BC and CACA are D,ED, E and FF respectively. If AB=8 cm,BC=10 cmAB = 8\text{ cm}, BC = 10\text{ cm} and CA=12 cmCA = 12\text{ cm}, find the perimeter of △DEF\triangle DEF.

Solution:

  1. According to the Mid-point Theorem, the segment joining the mid-points of two sides is half the third side.
  2. Therefore, DE=12AC=12×12=6 cmDE = \frac{1}{2} AC = \frac{1}{2} \times 12 = 6\text{ cm}.
  3. Similarly, EF=12AB=12×8=4 cmEF = \frac{1}{2} AB = \frac{1}{2} \times 8 = 4\text{ cm}.
  4. And DF=12BC=12×10=5 cmDF = \frac{1}{2} BC = \frac{1}{2} \times 10 = 5\text{ cm}.
  5. Perimeter of △DEF=DE+EF+DF=6+4+5=15 cm\triangle DEF = DE + EF + DF = 6 + 4 + 5 = 15\text{ cm}.

Explanation:

We use the Mid-point Theorem property where each side of the inner triangle is half the length of the side it is parallel to in the outer triangle.

Problem 2:

In △ABC\triangle ABC, ADAD is the median to BCBC. EE is the mid-point of ADAD. BEBE is produced to meet ACAC at FF. Prove that AF=13ACAF = \frac{1}{3} AC.

Solution:

  1. Draw DG∥BFDG \parallel BF meeting ACAC at GG.
  2. In △ADG\triangle ADG, EE is the mid-point of ADAD and EF∥DGEF \parallel DG. By the Converse of Mid-point Theorem, FF is the mid-point of AGAG. Thus, AF=FGAF = FG ... (i)
  3. In △BCF\triangle BCF, DD is the mid-point of BCBC (since ADAD is a median) and DG∥BFDG \parallel BF. By the Converse of Mid-point Theorem, GG is the mid-point of FCFC. Thus, FG=GCFG = GC ... (ii)
  4. From (i) and (ii), AF=FG=GCAF = FG = GC.
  5. Since AF+FG+GC=ACAF + FG + GC = AC, we have 3AF=AC3 AF = AC, which means AF=13ACAF = \frac{1}{3} AC.

Explanation:

This problem uses the Converse of the Mid-point Theorem twice. By constructing a parallel line DGDG, we create two triangles where the theorem can be applied to show that ACAC is divided into three equal segments.

Problem 3:

In the figure, M,NM, N and PP are the mid-points of AB,ACAB, AC and BCBC respectively. If MN=3 cm,NP=3.5 cmMN = 3\text{ cm}, NP = 3.5\text{ cm} and MP=2.5 cmMP = 2.5\text{ cm}, calculate the perimeter of △ABC\triangle ABC.

Triangle ABC with medial triangle MNP

Solution:

By the Mid-point Theorem:

  1. MN=12BC  ⟹  BC=2×MN=2×3=6 cmMN = \frac{1}{2} BC \implies BC = 2 \times MN = 2 \times 3 = 6\text{ cm}
  2. NP=12AB  ⟹  AB=2×NP=2×3.5=7 cmNP = \frac{1}{2} AB \implies AB = 2 \times NP = 2 \times 3.5 = 7\text{ cm}
  3. MP=12AC  ⟹  AC=2×MP=2×2.5=5 cmMP = \frac{1}{2} AC \implies AC = 2 \times MP = 2 \times 2.5 = 5\text{ cm} Perimeter of △ABC=AB+BC+AC\triangle ABC = AB + BC + AC Perimeter =7+6+5=18 cm= 7 + 6 + 5 = 18\text{ cm}

Explanation:

We use the Mid-point Theorem which states that a segment joining mid-points is half the length of the parallel side. We multiply each medial segment by 2 to find the lengths of the outer triangle's sides.

Problem 4:

In a trapezium ABCDABCD, AB∥DCAB \parallel DC. EE is the mid-point of ADAD. A line through EE parallel to ABAB meets BCBC at FF. Show that FF is the mid-point of BCBC.

Trapezium ABCD with mid-line EF and diagonal AC intersecting at G

Solution:

Join ACAC letting it intersect EFEF at GG. In △ADC\triangle ADC: EE is the mid-point of ADAD (given). EG∥DCEG \parallel DC (since EF∥ABEF \parallel AB and AB∥DCAB \parallel DC). By the Converse of Mid-point Theorem, GG must be the mid-point of ACAC. Now in △ABC\triangle ABC: GG is the mid-point of ACAC (proved above). GF∥ABGF \parallel AB (given EF∥ABEF \parallel AB). By the Converse of Mid-point Theorem, FF is the mid-point of BCBC.

Explanation:

To prove FF is a mid-point, we construct a diagonal to create two triangles. Applying the Converse of the Mid-point Theorem sequentially in both triangles confirms the result.