Quadrilaterals — NCERT Solutions, AI Tutor & Practice

The sum of the angles of a quadrilateral is $360^\circ$.

Let the angles be $3x, 5x, 9x$ and $13x$.

$3x + 5x + 9x + 13x = 360^\circ$ $30x = 360^\circ$ $x = 12^\circ$

So the angles are:

• $3x = 36^\circ$
• $5x = 60^\circ$
• $9x = 108^\circ$
• $13x = 156^\circ$

Check: $36 + 60 + 108 + 156 = 360^\circ$. ✓

Let $ABCD$ be a parallelogram with $AC = BD$. We prove it is a rectangle.

Consider triangles $ABC$ and $DCB$:

• $AB = DC$ (opposite sides of a parallelogram)
• $BC = CB$ (common)
• $AC = DB$ (given equal diagonals)

By SSS congruence, $\triangle ABC \cong \triangle DCB$.

Therefore $\angle ABC = \angle DCB$ (CPCT).

But $AB \parallel DC$, and $BC$ is a transversal, so these are co-interior angles: $\angle ABC + \angle DCB = 180^\circ$

Since the two angles are equal: $2\angle ABC = 180^\circ \Rightarrow \angle ABC = 90^\circ$

A parallelogram with one right angle is a rectangle. Hence $ABCD$ is a rectangle.

Let $ABCD$ be a rhombus with diagonals $AC$ and $BD$ meeting at $O$.

Since a rhombus is a parallelogram, its diagonals bisect each other: $OA = OC, \quad OB = OD$

Also all sides are equal, so $AB = AD$.

Consider triangles $AOB$ and $AOD$:

• $OB = OD$ (diagonals bisect each other)
• $AO = AO$ (common)
• $AB = AD$ (sides of a rhombus)

By SSS congruence, $\triangle AOB \cong \triangle AOD$.

Therefore $\angle AOB = \angle AOD$ (CPCT).

But $\angle AOB + \angle AOD = 180^\circ$ (linear pair on line $BD$).

So $2\angle AOB = 180^\circ \Rightarrow \angle AOB = 90^\circ$.

Thus the diagonals bisect each other at right angles.

In a parallelogram, opposite angles are equal and adjacent angles are supplementary.

Given $\angle A = 70^\circ$.

Opposite angle: $\angle C = \angle A = 70^\circ$.

Adjacent angle: $\angle A + \angle B = 180^\circ$ $\angle B = 180^\circ - 70^\circ = 110^\circ$

Opposite angle: $\angle D = \angle B = 110^\circ$.

So $\angle B = 110^\circ$, $\angle C = 70^\circ$, $\angle D = 110^\circ$.

In parallelogram $ABCD$, $AD \parallel BC$ with $AB$ as transversal, so adjacent angles are supplementary: $\angle A + \angle B = 180^\circ$

Divide by $2$: $\tfrac{1}{2}\angle A + \tfrac{1}{2}\angle B = 90^\circ$

Since $AP$ bisects $\angle A$ and $BP$ bisects $\angle B$: $\angle PAB = \tfrac{1}{2}\angle A, \qquad \angle PBA = \tfrac{1}{2}\angle B$

In $\triangle APB$, the sum of angles is $180^\circ$: $\angle PAB + \angle PBA + \angle APB = 180^\circ$ $90^\circ + \angle APB = 180^\circ$ $\angle APB = 90^\circ$

Given: $AB \parallel DC$ and $AD = BC$ (with $AB$ the longer parallel side).

Draw $DE \parallel CB$, where $E$ lies on $AB$.

Then $DCBE$ is a parallelogram (both pairs of opposite sides parallel: $DC \parallel EB$ and $DE \parallel CB$).

So $DE = CB = AD$ (given), making $\triangle ADE$ isosceles.

Therefore $\angle DAE = \angle DEA$ ... (i) (angles opposite equal sides).

Since $DE \parallel CB$, $\angle DEA = \angle CBE$ (corresponding angles) ... (ii)

From (i) and (ii): $\angle DAE = \angle CBE$, i.e. $\angle A = \angle B$.

Let $ABCD$ be a rectangle. By definition it is a parallelogram in which $\angle A = 90^\circ$.

Since opposite angles of a parallelogram are equal: $\angle C = \angle A = 90^\circ$

Since adjacent angles are supplementary: $\angle A + \angle B = 180^\circ \Rightarrow \angle B = 180^\circ - 90^\circ = 90^\circ$

And opposite to $\angle B$: $\angle D = \angle B = 90^\circ$

Hence $\angle A = \angle B = \angle C = \angle D = 90^\circ$, so each angle of a rectangle is a right angle.

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.