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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. Visually, if you have a triangle ABCABC and you mark point DD as the middle of ABAB and point EE as the middle of ACAC, the horizontal-looking line DEDE will be exactly parallel to the base BCBC and its length will be exactly BC/2BC/2.

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The Converse of the Mid-point Theorem states that a line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side. In a visual context, if you start at the mid-point DD of side ABAB and draw a line segment DEDE such that it is parallel to the base BCBC, it will always hit the side ACAC exactly at its middle point EE.

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The Mid-point Theorem is a specific case of the Basic Proportionality Theorem. While the Mid-point Theorem deals with a fixed ratio of 1:11:1, the broader concept shows that any line parallel to one side of a triangle divides the other two sides in the same ratio.

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When you join the mid-points of the four sides of any quadrilateral in order, the resulting inner figure is always a parallelogram. Visually, imagine an irregular four-sided shape ABCDABCD; if you connect the center points of AB,BC,CD,AB, BC, CD, and DADA, the resulting shape PQRSPQRS will have opposite sides that are parallel and equal in length.

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If there are three or more parallel lines and the intercepts made by them on one transversal are equal, then the intercepts made by them on any other transversal are also equal. This is known as the Intercept Theorem. Visually, if three parallel lines cut a line into two equal segments of 5extcm5 ext{ cm} each, they will cut any other slanted line into two segments that are equal to each other (though not necessarily 5extcm5 ext{ cm}).

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The triangle formed by joining the mid-points of the sides of a given triangle is called the Medial Triangle. The perimeter of this medial triangle is exactly half the perimeter of the original triangle because each of its sides is half the length of the corresponding side of the larger triangle.

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The four triangles formed by joining the mid-points of the sides of a triangle are congruent to each other and their areas are equal. Each small triangle has an area that is exactly 14\frac{1}{4} the area of the original large triangle.

📐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)

Area 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=8extcm,BC=10extcmAB = 8 ext{ cm}, BC = 10 ext{ cm} and CA=12extcmCA = 12 ext{ 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=6extcmDE = \frac{1}{2} AC = \frac{1}{2} \times 12 = 6 ext{ cm}.
  3. Similarly, EF=12AB=12×8=4extcmEF = \frac{1}{2} AB = \frac{1}{2} \times 8 = 4 ext{ cm}.
  4. And DF=12BC=12×10=5extcmDF = \frac{1}{2} BC = \frac{1}{2} \times 10 = 5 ext{ cm}.
  5. Perimeter of △DEF=DE+EF+DF=6+4+5=15extcm\triangle DEF = DE + EF + DF = 6 + 4 + 5 = 15 ext{ 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.