Exercise 8.2: Quadrilaterals — NCERT Solutions

Part 1 — D is the mid-point of AC.

In $\triangle ABC$, $M$ is the mid-point of $AB$ and $MD \parallel BC$.

By the converse of the mid-point theorem, a line through the mid-point of one side parallel to another side bisects the third side.

Therefore $D$ is the mid-point of $AC$.

Part 2 — $MD \perp AC$.

Since $MD \parallel BC$ and $\angle BCA = 90^\circ$ (right angle at $C$), the transversal $AC$ gives corresponding angles: $\angle MDA = \angle BCA = 90^\circ$

Hence $MD \perp AC$.

By the mid-point theorem, the segment joining the mid-points of two sides is parallel to and half of the third side:

• $EF \parallel BC$ and $EF = \tfrac{1}{2}BC = BD = DC$
• $DF \parallel AC$ and $DF = \tfrac{1}{2}AC = AE = EC$
• $DE \parallel AB$ and $DE = \tfrac{1}{2}AB = AF = FB$

Triangle $BDF$ vs $DEF$: $BDFE$-type reasoning — $DF = BD$? Instead compare directly:

In $\triangle AFE$ and $\triangle FBD$ and $\triangle EDC$ and $\triangle DEF$, all sides match:

• $\triangle AFE$: $AF = FB$, $AE = EC$, $FE = \tfrac12 BC = BD$
• Each small triangle has sides equal to $\tfrac12 AB,\ \tfrac12 BC,\ \tfrac12 CA$.

Hence by SSS congruence: $\triangle AFE \cong \triangle FBD \cong \triangle EDC \cong \triangle DEF$

Thus $\triangle DEF$ divides $\triangle ABC$ into four congruent triangles.

Let the diagonals meet at $O$.

Step 1: The diagonals bisect each other $\Rightarrow$ $ABCD$ is a parallelogram.

Step 2: The diagonals are equal $\Rightarrow$ the parallelogram is a rectangle (a parallelogram with equal diagonals is a rectangle).

Step 3: The diagonals meet at right angles $\Rightarrow$ the parallelogram is a rhombus (a parallelogram whose diagonals are perpendicular is a rhombus).

A quadrilateral that is both a rectangle and a rhombus is a square.

Hence $ABCD$ is a square.

Join the diagonal $AC$.

In $\triangle ABC$: $P$ and $Q$ are mid-points of $AB$ and $BC$. By the mid-point theorem: $PQ \parallel AC \quad\text{and}\quad PQ = \tfrac{1}{2}AC$

In $\triangle ADC$: $S$ and $R$ are mid-points of $AD$ and $CD$. By the mid-point theorem: $SR \parallel AC \quad\text{and}\quad SR = \tfrac{1}{2}AC$

Therefore: $PQ \parallel SR \quad\text{and}\quad PQ = SR$

A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.

Hence $PQRS$ is a parallelogram.