Quadrilaterals
NCERT Solutions • Class 9 Maths • Chapter 8Exercise 8.2
1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that:
(i) $SR \parallel AC$ and $SR = \frac{1}{2}AC$
(ii) $PQ = SR$
(iii) $PQRS$ is a parallelogram.
(i) $SR \parallel AC$ and $SR = \frac{1}{2}AC$
(ii) $PQ = SR$
(iii) $PQRS$ is a parallelogram.
(i) In $\Delta ADC$:
$S$ is mid-point of $DA$ and $R$ is mid-point of $CD$.
By Mid-point Theorem:
$SR \parallel AC$ and $SR = \frac{1}{2}AC$. …(1)
$S$ is mid-point of $DA$ and $R$ is mid-point of $CD$.
By Mid-point Theorem:
$SR \parallel AC$ and $SR = \frac{1}{2}AC$. …(1)
(ii) In $\Delta ABC$:
$P$ is mid-point of $AB$ and $Q$ is mid-point of $BC$.
By Mid-point Theorem:
$PQ \parallel AC$ and $PQ = \frac{1}{2}AC$. …(2)
From (1) and (2):
$PQ = SR$.
$P$ is mid-point of $AB$ and $Q$ is mid-point of $BC$.
By Mid-point Theorem:
$PQ \parallel AC$ and $PQ = \frac{1}{2}AC$. …(2)
From (1) and (2):
$PQ = SR$.
(iii) From (1) and (2):
$PQ \parallel AC$ and $SR \parallel AC \Rightarrow PQ \parallel SR$.
Also, $PQ = SR$.
Since one pair of opposite sides is equal and parallel:
$PQRS$ is a parallelogram.
$PQ \parallel AC$ and $SR \parallel AC \Rightarrow PQ \parallel SR$.
Also, $PQ = SR$.
Since one pair of opposite sides is equal and parallel:
$PQRS$ is a parallelogram.
2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Step 1: Prove $PQRS$ is a parallelogram.
Using the logic from Q1, $PQ \parallel AC$ and $SR \parallel AC \Rightarrow PQ \parallel SR$ and $PQ = SR$. Thus, $PQRS$ is a parallelogram.
Using the logic from Q1, $PQ \parallel AC$ and $SR \parallel AC \Rightarrow PQ \parallel SR$ and $PQ = SR$. Thus, $PQRS$ is a parallelogram.
Step 2: Prove one angle is $90^\circ$.
Diagonals of a rhombus intersect at $90^\circ$. So, $AC \perp BD$.
Since $PQ \parallel AC$ and $QR \parallel BD$ (by Mid-point theorem on $\Delta BCD$),
The angle between $PQ$ and $QR$ equals the angle between $AC$ and $BD$.
$\therefore \angle PQR = 90^\circ$.
A parallelogram with one right angle is a rectangle.
Diagonals of a rhombus intersect at $90^\circ$. So, $AC \perp BD$.
Since $PQ \parallel AC$ and $QR \parallel BD$ (by Mid-point theorem on $\Delta BCD$),
The angle between $PQ$ and $QR$ equals the angle between $AC$ and $BD$.
$\therefore \angle PQR = 90^\circ$.
A parallelogram with one right angle is a rectangle.
3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Step 1: $PQRS$ is a parallelogram (from Q1 logic).
Step 2: Show adjacent sides are equal.
In rectangle $ABCD$, diagonal $AC = BD$.
From mid-point theorem:
$PQ = \frac{1}{2} AC$ and $QR = \frac{1}{2} BD$.
Since $AC = BD$, it implies $PQ = QR$.
A parallelogram with equal adjacent sides is a rhombus.
In rectangle $ABCD$, diagonal $AC = BD$.
From mid-point theorem:
$PQ = \frac{1}{2} AC$ and $QR = \frac{1}{2} BD$.
Since $AC = BD$, it implies $PQ = QR$.
A parallelogram with equal adjacent sides is a rhombus.
4. ABCD is a trapezium in which $AB \parallel DC$, $BD$ is a diagonal and $E$ is the mid-point of $AD$. A line is drawn through $E$ parallel to $AB$ intersecting $BC$ at $F$. Show that $F$ is the mid-point of $BC$.
Let $BD$ intersect $EF$ at $G$.
In $\Delta DAB$:
$E$ is mid-point of $AD$ and $EG \parallel AB$.
By Converse of Mid-point Theorem, $G$ is the mid-point of $BD$.
In $\Delta DAB$:
$E$ is mid-point of $AD$ and $EG \parallel AB$.
By Converse of Mid-point Theorem, $G$ is the mid-point of $BD$.
In $\Delta BCD$:
$G$ is the mid-point of $BD$.
Since $EF \parallel AB$ and $AB \parallel DC$, then $GF \parallel DC$.
By Converse of Mid-point Theorem, $F$ is the mid-point of $BC$.
$G$ is the mid-point of $BD$.
Since $EF \parallel AB$ and $AB \parallel DC$, then $GF \parallel DC$.
By Converse of Mid-point Theorem, $F$ is the mid-point of $BC$.
5. In a parallelogram $ABCD$, $E$ and $F$ are the mid-points of sides $AB$ and $CD$ respectively. Show that the line segments $AF$ and $EC$ trisect the diagonal $BD$.
Step 1: Show $AECF$ is a parallelogram.
$AB \parallel CD \Rightarrow AE \parallel CF$.
$AB = CD \Rightarrow \frac{1}{2}AB = \frac{1}{2}CD \Rightarrow AE = CF$.
Thus, $AECF$ is a parallelogram $\Rightarrow AF \parallel EC$.
$AB \parallel CD \Rightarrow AE \parallel CF$.
$AB = CD \Rightarrow \frac{1}{2}AB = \frac{1}{2}CD \Rightarrow AE = CF$.
Thus, $AECF$ is a parallelogram $\Rightarrow AF \parallel EC$.
Step 2: In $\Delta DQC$ (Let $P$ and $Q$ be intersection points on $BD$):
$F$ is mid-point of $CD$ and $FP \parallel CQ$.
$\therefore P$ is mid-point of $DQ \Rightarrow DP = PQ$. …(1)
$F$ is mid-point of $CD$ and $FP \parallel CQ$.
$\therefore P$ is mid-point of $DQ \Rightarrow DP = PQ$. …(1)
Step 3: In $\Delta APB$:
$E$ is mid-point of $AB$ and $EQ \parallel AP$.
$\therefore Q$ is mid-point of $PB \Rightarrow PQ = QB$. …(2)
From (1) and (2):
$DP = PQ = QB$. Hence, trisected.
$E$ is mid-point of $AB$ and $EQ \parallel AP$.
$\therefore Q$ is mid-point of $PB \Rightarrow PQ = QB$. …(2)
From (1) and (2):
$DP = PQ = QB$. Hence, trisected.
6. $ABC$ is a triangle right angled at $C$. A line through the mid-point $M$ of hypotenuse $AB$ and parallel to $BC$ intersects $AC$ at $D$. Show that:
(i) $D$ is the mid-point of $AC$
(ii) $MD \perp AC$
(iii) $CM = MA = \frac{1}{2} AB$
(i) $D$ is the mid-point of $AC$
(ii) $MD \perp AC$
(iii) $CM = MA = \frac{1}{2} AB$
(i) In $\Delta ABC$:
$M$ is mid-point of $AB$ and $MD \parallel BC$.
By Converse of Mid-point Theorem, $D$ is the mid-point of $AC$.
$M$ is mid-point of $AB$ and $MD \parallel BC$.
By Converse of Mid-point Theorem, $D$ is the mid-point of $AC$.
(ii) Since $MD \parallel BC$, corresponding angles are equal:
$\angle ADM = \angle ACB = 90^\circ$.
Therefore, $MD \perp AC$.
$\angle ADM = \angle ACB = 90^\circ$.
Therefore, $MD \perp AC$.
(iii) In $\Delta ADM$ and $\Delta CDM$:
1. $AD = CD$ (from i)
2. $\angle ADM = \angle CDM = 90^\circ$ (from ii)
3. $DM = DM$ (Common)
$\therefore \Delta ADM \cong \Delta CDM$ (SAS).
$\Rightarrow AM = CM$ (CPCT).
Since $AM = \frac{1}{2} AB$, then $CM = MA = \frac{1}{2} AB$.
1. $AD = CD$ (from i)
2. $\angle ADM = \angle CDM = 90^\circ$ (from ii)
3. $DM = DM$ (Common)
$\therefore \Delta ADM \cong \Delta CDM$ (SAS).
$\Rightarrow AM = CM$ (CPCT).
Since $AM = \frac{1}{2} AB$, then $CM = MA = \frac{1}{2} AB$.