Quadrilaterals
NCERT Solutions • Class 9 Maths • Chapter 8Exercise 8.1
1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Given: $ABCD$ is a parallelogram where $AC = BD$.
Proof: In $\Delta ABC$ and $\Delta DCB$:
1. $AB = DC$ (Opposite sides of parallelogram)
2. $BC = CB$ (Common)
3. $AC = DB$ (Given)
$\therefore \Delta ABC \cong \Delta DCB$ (By SSS Rule).
$\Rightarrow \angle ABC = \angle DCB$ (CPCT).
Also, $\angle ABC + \angle DCB = 180^\circ$ (Consecutive interior angles).
$\Rightarrow 2\angle ABC = 180^\circ \Rightarrow \angle ABC = 90^\circ$.
Since one angle is $90^\circ$, ABCD is a rectangle.
Proof: In $\Delta ABC$ and $\Delta DCB$:
1. $AB = DC$ (Opposite sides of parallelogram)
2. $BC = CB$ (Common)
3. $AC = DB$ (Given)
$\therefore \Delta ABC \cong \Delta DCB$ (By SSS Rule).
$\Rightarrow \angle ABC = \angle DCB$ (CPCT).
Also, $\angle ABC + \angle DCB = 180^\circ$ (Consecutive interior angles).
$\Rightarrow 2\angle ABC = 180^\circ \Rightarrow \angle ABC = 90^\circ$.
Since one angle is $90^\circ$, ABCD is a rectangle.
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2. Show that the diagonals of a square are equal and bisect each other at right angles.
Part 1 (Equal): In $\Delta ABC$ and $\Delta BAD$:
$AB=AB$ (Common), $BC=AD$ (Sides of square), $\angle ABC = \angle BAD = 90^\circ$.
$\therefore \Delta ABC \cong \Delta BAD \Rightarrow$ $AC = BD$.
$AB=AB$ (Common), $BC=AD$ (Sides of square), $\angle ABC = \angle BAD = 90^\circ$.
$\therefore \Delta ABC \cong \Delta BAD \Rightarrow$ $AC = BD$.
Part 2 (Bisect): Since a square is a parallelogram, diagonals bisect each other.
$\Rightarrow OA = OC$ and $OB = OD$.
$\Rightarrow OA = OC$ and $OB = OD$.
Part 3 (Right Angles): In $\Delta AOB$ and $\Delta AOD$:
$OB = OD$ (Diagonals bisect), $AB = AD$ (Sides of square), $AO = AO$ (Common).
$\therefore \Delta AOB \cong \Delta AOD \Rightarrow \angle AOB = \angle AOD$.
Also, $\angle AOB + \angle AOD = 180^\circ$ (Linear Pair).
$\Rightarrow 2\angle AOB = 180^\circ \Rightarrow$ $\angle AOB = 90^\circ$.
$OB = OD$ (Diagonals bisect), $AB = AD$ (Sides of square), $AO = AO$ (Common).
$\therefore \Delta AOB \cong \Delta AOD \Rightarrow \angle AOB = \angle AOD$.
Also, $\angle AOB + \angle AOD = 180^\circ$ (Linear Pair).
$\Rightarrow 2\angle AOB = 180^\circ \Rightarrow$ $\angle AOB = 90^\circ$.
3. Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle A$. Show that:
(i) it bisects $\angle C$ also.
(ii) $ABCD$ is a rhombus.
(i) it bisects $\angle C$ also.
(ii) $ABCD$ is a rhombus.
(i) Since $AB \parallel DC$, $\angle 1 = \angle 4$ (Alternate Interior Angles).
Since $AD \parallel BC$, $\angle 2 = \angle 3$ (Alternate Interior Angles).
Given $\angle 1 = \angle 2$ (AC bisects A).
$\therefore \angle 3 = \angle 4$. Hence, $AC$ bisects $\angle C$.
Since $AD \parallel BC$, $\angle 2 = \angle 3$ (Alternate Interior Angles).
Given $\angle 1 = \angle 2$ (AC bisects A).
$\therefore \angle 3 = \angle 4$. Hence, $AC$ bisects $\angle C$.
(ii) From above, $\angle 1 = \angle 2$ and $\angle 2 = \angle 3$, so $\angle 1 = \angle 3$.
In $\Delta ADC$, sides opposite to equal angles are equal $\Rightarrow AD = CD$.
Since adjacent sides are equal in a parallelogram, $ABCD$ is a rhombus.
In $\Delta ADC$, sides opposite to equal angles are equal $\Rightarrow AD = CD$.
Since adjacent sides are equal in a parallelogram, $ABCD$ is a rhombus.
4. $ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ as well as $\angle C$. Show that:
(i) $ABCD$ is a square.
(ii) Diagonal $BD$ bisects $\angle B$ as well as $\angle D$.
(i) $ABCD$ is a square.
(ii) Diagonal $BD$ bisects $\angle B$ as well as $\angle D$.
(i) Since $AC$ bisects $\angle A$ and $\angle C$, $\angle DAC = \angle BAC$.
Also, $\angle DAC = \angle BCA$ (Alternate Angles).
$\Rightarrow \angle BAC = \angle BCA$.
In $\Delta ABC$, $AB = BC$ (Sides opposite equal angles).
A rectangle with equal adjacent sides is a square.
Also, $\angle DAC = \angle BCA$ (Alternate Angles).
$\Rightarrow \angle BAC = \angle BCA$.
In $\Delta ABC$, $AB = BC$ (Sides opposite equal angles).
A rectangle with equal adjacent sides is a square.
(ii) Since $ABCD$ is a square, diagonals bisect the angles.
In $\Delta BCD$, $BC = CD \Rightarrow \angle CBD = \angle CDB$.
Also, alternate angles are equal.
Therefore, $BD$ bisects $\angle B$ and $\angle D$.
In $\Delta BCD$, $BC = CD \Rightarrow \angle CBD = \angle CDB$.
Also, alternate angles are equal.
Therefore, $BD$ bisects $\angle B$ and $\angle D$.
5. In parallelogram $ABCD$, points $P$ and $Q$ are on $BD$ such that $DP = BQ$. Show that:
(i) $\Delta APD \cong \Delta CQB$
(ii) $AP = CQ$
(iii) $\Delta AQB \cong \Delta CPD$
(iv) $AQ = CP$
(v) $APCQ$ is a parallelogram
(i) $\Delta APD \cong \Delta CQB$
(ii) $AP = CQ$
(iii) $\Delta AQB \cong \Delta CPD$
(iv) $AQ = CP$
(v) $APCQ$ is a parallelogram
(i) In $\Delta APD$ and $\Delta CQB$:
1. $AD = CB$ (Opposite sides)
2. $\angle ADP = \angle CBQ$ (Alt. interior angles)
3. $DP = BQ$ (Given)
$\therefore$ $\Delta APD \cong \Delta CQB$ (SAS).
1. $AD = CB$ (Opposite sides)
2. $\angle ADP = \angle CBQ$ (Alt. interior angles)
3. $DP = BQ$ (Given)
$\therefore$ $\Delta APD \cong \Delta CQB$ (SAS).
(ii) From (i), by CPCT: $AP = CQ$.
(iii) In $\Delta AQB$ and $\Delta CPD$:
1. $AB = CD$ (Opposite sides)
2. $\angle ABQ = \angle CDP$ (Alt. interior angles)
3. $BQ = DP$ (Given)
$\therefore$ $\Delta AQB \cong \Delta CPD$ (SAS).
1. $AB = CD$ (Opposite sides)
2. $\angle ABQ = \angle CDP$ (Alt. interior angles)
3. $BQ = DP$ (Given)
$\therefore$ $\Delta AQB \cong \Delta CPD$ (SAS).
(iv) From (iii), by CPCT: $AQ = CP$.
(v) Since $AP = CQ$ and $AQ = CP$ (Opposite sides equal):
$APCQ$ is a parallelogram.
$APCQ$ is a parallelogram.
6. $ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from $A$ and $C$ on $BD$. Show that:
(i) $\Delta APB \cong \Delta CQD$
(ii) $AP = CQ$
(i) $\Delta APB \cong \Delta CQD$
(ii) $AP = CQ$
(i) In $\Delta APB$ and $\Delta CQD$:
1. $\angle APB = \angle CQD = 90^\circ$ (Perpendiculars)
2. $\angle ABP = \angle CDQ$ (Alt. interior angles)
3. $AB = CD$ (Opposite sides)
$\therefore$ $\Delta APB \cong \Delta CQD$ (By AAS Rule).
1. $\angle APB = \angle CQD = 90^\circ$ (Perpendiculars)
2. $\angle ABP = \angle CDQ$ (Alt. interior angles)
3. $AB = CD$ (Opposite sides)
$\therefore$ $\Delta APB \cong \Delta CQD$ (By AAS Rule).
(ii) From (i), by CPCT: $AP = CQ$.
7. $ABCD$ is a trapezium in which $AB \parallel CD$ and $AD = BC$. Show that:
(i) $\angle A = \angle B$
(ii) $\angle C = \angle D$
(iii) $\Delta ABC \cong \Delta BAD$
(iv) Diagonal $AC =$ Diagonal $BD$
(i) $\angle A = \angle B$
(ii) $\angle C = \angle D$
(iii) $\Delta ABC \cong \Delta BAD$
(iv) Diagonal $AC =$ Diagonal $BD$
Construction: Draw a line through $C$ parallel to $DA$ intersecting $AB$ produced at $E$.
$ADCE$ is a parallelogram $\Rightarrow AD = CE$. But $AD = BC$, so $BC = CE$.
In $\Delta BCE$, $\angle CBE = \angle CEB$.
$ADCE$ is a parallelogram $\Rightarrow AD = CE$. But $AD = BC$, so $BC = CE$.
In $\Delta BCE$, $\angle CBE = \angle CEB$.
(i) $\angle A + \angle CEB = 180^\circ$ (Interior angles).
$\angle B + \angle CBE = 180^\circ$ (Linear pair).
Since $\angle CBE = \angle CEB$, we get $\angle A = \angle B$.
$\angle B + \angle CBE = 180^\circ$ (Linear pair).
Since $\angle CBE = \angle CEB$, we get $\angle A = \angle B$.
(ii) In trapezium $ABCD$, $\angle A + \angle D = 180^\circ$ and $\angle B + \angle C = 180^\circ$.
Since $\angle A = \angle B$, then $\angle C = \angle D$.
Since $\angle A = \angle B$, then $\angle C = \angle D$.
(iii) In $\Delta ABC$ and $\Delta BAD$:
1. $AB = BA$ (Common)
2. $\angle B = \angle A$ (Proved)
3. $BC = AD$ (Given)
$\therefore$ $\Delta ABC \cong \Delta BAD$ (SAS).
1. $AB = BA$ (Common)
2. $\angle B = \angle A$ (Proved)
3. $BC = AD$ (Given)
$\therefore$ $\Delta ABC \cong \Delta BAD$ (SAS).
(iv) From (iii), by CPCT: $AC = BD$.