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Quadrilaterals - Applications of Parallelograms

Grade 9CBSE

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 two sides of a triangle is parallel to the third side and is equal to half of it. In △ABC\triangle ABC, if DD and EE are mid-points of ABAB and ACAC, then DE∥BCDE \parallel BC and DE=12BCDE = \frac{1}{2}BC.

Triangle ABC with midpoints D and E joined by segment DE
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The Converse of 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.

A line drawn from midpoint D parallel to the base BC
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A quadrilateral formed by joining the mid-points of the sides of any quadrilateral, in order, is a parallelogram. This is a direct application of the Mid-point Theorem using diagonals.

A quadrilateral with another quadrilateral formed by its midpoints inside

📐Formulae

DE∥BC and DE=12BCDE \parallel BC \text{ and } DE = \frac{1}{2}BC

In △ABC, if AD=DB and AE=EC⇒DE=12BC\text{In } \triangle ABC, \text{ if } AD = DB \text{ and } AE = EC \Rightarrow DE = \frac{1}{2}BC

💡Examples

Problem 1:

In △ABC\triangle ABC, DD, EE and FF are respectively the mid-points of sides ABAB, BCBC and CACA. Show that △ABC\triangle ABC is divided into four congruent triangles by joining DD, EE and FF.

Triangle ABC with midpoints D, E, F forming four internal triangles

Solution:

  1. Since DD and FF are mid-points of ABAB and ACAC, by Mid-point Theorem, DF∥BCDF \parallel BC and DF=12BC=BEDF = \frac{1}{2}BC = BE.
  2. Thus, BDEFBDEF is a parallelogram. Similarly, DCEFDCEF and DEAFDEAF are parallelograms.
  3. A diagonal of a parallelogram divides it into two congruent triangles.
  4. △BDE≅△FED\triangle BDE \cong \triangle FED (diagonal DEDE of parallelogram BDEFBDEF)
  5. △CEF≅△FED\triangle CEF \cong \triangle FED (diagonal EFEF of parallelogram DCEFDCEF)
  6. △ADF≅△FED\triangle ADF \cong \triangle FED (diagonal DFDF of parallelogram DEAFDEAF)
  7. Therefore, all four triangles are congruent.

Explanation:

This problem uses the Mid-point theorem to establish that the smaller triangles form parallelograms, then uses the property that diagonals of a parallelogram bisect it into congruent triangles.

Problem 2:

ABCDABCD is a quadrilateral in which PP, QQ, RR and SS are mid-points of the sides ABAB, BCBC, CDCD and DADA. ACAC is a diagonal. Show that SR∥ACSR \parallel AC and SR=12ACSR = \frac{1}{2}AC.

Quadrilateral ABCD with diagonal AC and midpoints S and R

Solution:

  1. Consider △ADC\triangle ADC. SS is the mid-point of ADAD and RR is the mid-point of CDCD.
  2. By the Mid-point Theorem, in a triangle, the line segment joining the mid-points of two sides is parallel to the third side and half of it.
  3. Therefore, SR∥ACSR \parallel AC and SR=12ACSR = \frac{1}{2}AC.
  4. Similarly, in △ABC\triangle ABC, PQ∥ACPQ \parallel AC and PQ=12ACPQ = \frac{1}{2}AC.
  5. From these, we can conclude PQ∥SRPQ \parallel SR and PQ=SRPQ = SR.

Explanation:

Applying the mid-point theorem to triangles formed by the diagonal of a quadrilateral proves the relationship between the mid-point segment and the diagonal.