krit.club logo

Quadrilaterals - The Mid-point Theorem

Grade 9CBSE

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

🔑Concepts

•

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

•

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. If you start at the mid-point PP of side XYXY and draw a line parallel to base YZYZ, it will hit the side XZXZ exactly at its mid-point QQ.

•

The quadrilateral formed by joining the mid-points of the sides of any quadrilateral, in order, is a parallelogram. In a quadrilateral ABCDABCD with mid-points P,Q,R,SP, Q, R, S on sides AB,BC,CD,DAAB, BC, CD, DA, the segment PQPQ will be parallel to diagonal ACAC, and RSRS will also be parallel to ACAC. This makes PQ∥RSPQ \parallel RS, which is a defining property of a parallelogram.

•

If the mid-points of the sides of a rectangle are joined in order, the resulting quadrilateral is a rhombus. This occurs because the diagonals of a rectangle are equal in length, meaning all four sides of the inscribed quadrilateral (each being half the diagonal length) will be equal.

•

If the mid-points of the sides of a rhombus are joined in order, the resulting quadrilateral is a rectangle. Visually, because the diagonals of a rhombus are perpendicular, the adjacent sides of the inscribed quadrilateral will meet at 90∘90^{\circ} angles.

•

The perimeter of the triangle formed by joining the mid-points of the sides of △ABC\triangle ABC is equal to half the perimeter of △ABC\triangle ABC. Each side of the smaller interior triangle is exactly half the length of the corresponding outer side it is parallel to.

•

In a trapezoid, the segment joining the mid-points of the non-parallel sides is parallel to the parallel bases and its length is the average (half the sum) of the lengths of the two bases. If ABCDABCD is a trapezoid with AB∥CDAB \parallel CD, and E,FE, F are mid-points of ADAD and BCBC, then EF=12(AB+CD)EF = \frac{1}{2}(AB + CD).

📐Formulae

If D,ED, E are mid-points of sides AB,ACAB, AC in △ABC\triangle ABC, then: DE∥BCDE \parallel BC

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

In △ABC\triangle ABC, if DD is mid-point of ABAB and DE∥BCDE \parallel BC, then AE=ECAE = EC.

Perimeter(△DEF)=12×Perimeter(△ABC)\text{Perimeter}(\triangle DEF) = \frac{1}{2} \times \text{Perimeter}(\triangle ABC) where D,E,FD, E, F are mid-points.

For quadrilateral ABCDABCD with mid-points P,Q,R,SP, Q, R, S, the sides of parallelogram PQRSPQRS are: PQ=SR=12ACPQ = SR = \frac{1}{2} AC and QR=PS=12BDQR = PS = \frac{1}{2} BD.

💡Examples

Problem 1:

In △ABC\triangle ABC, D,ED, E and FF are the mid-points of sides AB,BCAB, BC and CACA respectively. If AB=6 cmAB = 6 \text{ cm}, BC=7 cmBC = 7 \text{ cm} and AC=8 cmAC = 8 \text{ cm}, find the perimeter of △DEF\triangle DEF.

Solution:

  1. By the Mid-point Theorem, DEDE joins mid-points of ABAB and BCBC, so DE=12AC=12×8=4 cmDE = \frac{1}{2} AC = \frac{1}{2} \times 8 = 4 \text{ cm}.
  2. Similarly, EFEF joins mid-points of BCBC and CACA, so EF=12AB=12×6=3 cmEF = \frac{1}{2} AB = \frac{1}{2} \times 6 = 3 \text{ cm}.
  3. DFDF joins mid-points of ABAB and CACA, so DF=12BC=12×7=3.5 cmDF = \frac{1}{2} BC = \frac{1}{2} \times 7 = 3.5 \text{ cm}.
  4. Perimeter of △DEF=DE+EF+DF=4+3+3.5=10.5 cm\triangle DEF = DE + EF + DF = 4 + 3 + 3.5 = 10.5 \text{ cm}.

Explanation:

This approach uses the Mid-point Theorem to determine each side of the inner triangle by taking half the length of the side of the original triangle to which it is parallel.

Problem 2:

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

Solution:

  1. Draw a line DG∥BFDG \parallel BF meeting ACAC at GG.
  2. 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.
  3. In △ADG\triangle ADG, EE is the mid-point of ADAD and EF∥DGEF \parallel DG (as EFEF is part of BFBF). By the Converse of Mid-point Theorem, FF is the mid-point of AGAG. Thus, AF=FGAF = FG.
  4. Combining the results: AF=FG=GCAF = FG = GC. Since AF+FG+GC=ACAF + FG + GC = AC, we have 3×AF=AC3 \times AF = AC.
  5. Therefore, AF=13ACAF = \frac{1}{3} AC.

Explanation:

This solution utilizes a construction to create a new triangle where the Converse of the Mid-point Theorem can be applied twice to show that the segments on side ACAC are equal.