Quadrilaterals — NCERT Solutions

Given: ABCD is a quadrilateral; P, Q, R, S are mid-points of AB, BC, CD, DA respectively.

(i) $SR \parallel AC$ and $SR = \frac{1}{2}AC$:

In $\triangle DAC$:

• S is the mid-point of DA and R is the mid-point of DC.

By the Mid-point Theorem:
$$SR \parallel AC \quad \text{and} \quad SR = \frac{1}{2}AC \checkmark$$

(ii) $PQ = SR$:

In $\triangle BAC$:

• P is the mid-point of AB and Q is the mid-point of BC.

By the Mid-point Theorem:
$$PQ \parallel AC \quad \text{and} \quad PQ = \frac{1}{2}AC$$

From (i): $SR = \frac{1}{2}AC$

$$\therefore PQ = SR \checkmark$$

(iii) PQRS is a parallelogram:

From (i) and (ii):
$$PQ \parallel AC \parallel SR \Rightarrow PQ \parallel SR$$
$$PQ = SR = \frac{1}{2}AC$$

Since one pair of opposite sides (PQ and SR) is both equal and parallel, PQRS is a parallelogram. $\blacksquare$

Given: ABCD is a rhombus; P, Q, R, S are mid-points of AB, BC, CD, DA respectively.

To prove: PQRS is a rectangle.

Construction: Join diagonals AC and BD.

Step 1: PQRS is a parallelogram.

In $\triangle ABC$: P and Q are mid-points of AB and BC.
$$\Rightarrow PQ \parallel AC \text{ and } PQ = \frac{1}{2}AC \quad \text{(Mid-point Theorem)}$$

In $\triangle ADC$: S and R are mid-points of AD and DC.
$$\Rightarrow SR \parallel AC \text{ and } SR = \frac{1}{2}AC$$

$\therefore PQ \parallel SR$ and $PQ = SR$, so PQRS is a parallelogram.

Step 2: One angle of PQRS is $90^\circ$.

In $\triangle ABD$: P and S are mid-points of AB and AD.
$$\Rightarrow PS \parallel BD$$

Now, the diagonals of a rhombus bisect each other at right angles, so $AC \perp BD$.

Since $PQ \parallel AC$ and $PS \parallel BD$, and $AC \perp BD$:
$$\angle QPS = 90^\circ$$

Since PQRS is a parallelogram with one angle $= 90^\circ$, PQRS is a rectangle. $\blacksquare$

Given: ABCD is a rectangle; P, Q, R, S are mid-points of AB, BC, CD, DA respectively.

To prove: PQRS is a rhombus.

Construction: Join diagonals AC and BD.

Step 1: PQRS is a parallelogram.

In $\triangle ABC$: P, Q are mid-points of AB, BC.
$$PQ \parallel AC,\; PQ = \frac{1}{2}AC$$

In $\triangle ADC$: S, R are mid-points of AD, DC.
$$SR \parallel AC,\; SR = \frac{1}{2}AC$$

$\therefore PQ \parallel SR$ and $PQ = SR$ $\Rightarrow$ PQRS is a parallelogram.

Step 2: Adjacent sides of PQRS are equal.

Since ABCD is a rectangle, $AC = BD$ (diagonals of a rectangle are equal).

In $\triangle ABD$: P, S are mid-points of AB, AD.
$$PS = \frac{1}{2}BD = \frac{1}{2}AC = PQ$$

Since adjacent sides $PQ = PS$, and PQRS is a parallelogram, all four sides are equal.

Hence PQRS is a rhombus. $\blacksquare$

Given: ABCD is a trapezium with $AB \parallel DC$; E is the mid-point of AD; $EF \parallel AB$, where F lies on BC.

To prove: F is the mid-point of BC.

Proof:

Let diagonal BD intersect EF at G.

In $\triangle ABD$:

• E is the mid-point of AD (given)
• $EG \parallel AB$ (since $EF \parallel AB$)

By the Converse of Mid-point Theorem, G is the mid-point of BD.

Now in $\triangle BDC$:

• G is the mid-point of BD (proved above)
• $GF \parallel DC$ (since $AB \parallel DC$ and $EF \parallel AB \Rightarrow EF \parallel DC$)

By the Converse of Mid-point Theorem, F is the mid-point of BC. $\blacksquare$

Given: ABCD is a parallelogram; E is the mid-point of AB and F is the mid-point of CD.

To prove: AF and EC trisect diagonal BD, i.e., $BP = PQ = QD$ where P and Q are the intersections of AF and EC with BD respectively.

Proof:

Since ABCD is a parallelogram:
$$AB \parallel CD \quad \text{and} \quad AB = CD$$

Since E and F are mid-points:
$$AE = \frac{1}{2}AB \quad \text{and} \quad CF = \frac{1}{2}CD$$

$$\Rightarrow AE = CF \quad \text{and} \quad AE \parallel CF$$

Therefore AECF is a parallelogram (one pair of opposite sides equal and parallel).

$\Rightarrow AF \parallel EC$ (opposite sides of parallelogram AECF).

Let AF intersect BD at P and EC intersect BD at Q.

In $\triangle ABP$: Since $AE \parallel QC$ ...

Consider $\triangle BPQ$ and use the following approach:

In $\triangle ABQ$ (where Q is on BD):

• $E$ is the mid-point of $AB$
• $EP \parallel AQ$ (since $AF \parallel EC$, so $EP \parallel BQ$... )

Let us use a cleaner approach:

In $\triangle DCQ$: F is the mid-point of DC and $FQ \parallel$ ...

Cleaner proof:

Since $AE \parallel FC$ and $AE = FC$, AECF is a parallelogram, so $AF \parallel EC$.

In $\triangle ABP$ (P = intersection of AF and BD):
Consider $\triangle DQC$ (Q = intersection of EC and BD):

In $\triangle ABP$: $EP \parallel$ ...

Actually, consider $\triangle BEP$ and the line through E:

In $\triangle ABD$:

• E is the mid-point of AB
• $EP \parallel AD$ (since $AEFD$ ... $AF \parallel EC$ and $AE \parallel FC$, so in $\triangle ABD$, $EP \parallel AD$)

Wait — since $AF \parallel EC$ means $AP \parallel EQ$... Let us use the correct argument:

Since ABCD is a parallelogram, $AB \parallel DC$, so $EB \parallel FC$. Also $EB = \frac{1}{2}AB = \frac{1}{2}DC = FC$. So EBCF is also a parallelogram, giving $EF \parallel BC$ and $EF = BC$.

Now in $\triangle ABD$: E is mid-point of AB and $EP \parallel AD$ (since $EP$ is part of $AF$, and $AF$... )

Let us use the most direct method:

In $\triangle ABD$, consider point P on BD where AF meets BD.
Since $AE \parallel FC$ (both parallel to $\frac{1}{2}$ of the same side), and AEFD is ...

Note: $AF$ is a diagonal of parallelogram AEFD (since $AE \parallel DF$ and $AE = DF = \frac{1}{2}AB = \frac{1}{2}DC$). Wait, $DF = DC - FC = DC - \frac{1}{2}DC = \frac{1}{2}DC = \frac{1}{2}AB = AE$. And $AE \parallel DF$ (both parallel to AB/DC). So AEFD is a parallelogram.

In $\triangle ABD$:

• E is mid-point of AB
• $EP \parallel AD$ (since in parallelogram AEFD, $EF \parallel AD$, and P lies on EF... no, P lies on AF)

Final clean proof:

In $\triangle ABP$ where P is on BD:
Since $AEFD$ is a parallelogram (shown above), $EF \parallel AD$.
In $\triangle ADB$: $E$ is mid-point of $AB$ and $EP \parallel AD$ (as $EF \parallel AD$ and P is on line $AF$).
By converse of mid-point theorem, P is mid-point of $DB$... that would give $BP = PD$, not trisection.

Correct approach — using two applications:

In $\triangle ABD$: $E$ is mid-point of $AB$. Line $EP$ where $EP \parallel AD$ (since $AE \parallel DF$ means $EP \parallel DP$... ).

Let me restart with the standard textbook proof:

Since $AE \parallel FC$ and $AE = FC$, AECF is a parallelogram $\Rightarrow$ $EC \parallel AF$.

In $\triangle ABQ$: $E$ is mid-point of $AB$ and $EP \parallel BQ$ (since $AF \parallel EC$ means the segment from $E$ parallel to $AQ$)...

Standard solution:

Since $ABCD$ is a parallelogram, $AB = DC$ and $AB \parallel DC$.
$E$ is mid-point of $AB$ and $F$ is mid-point of $DC$.
$\therefore AE = EB = DF = FC = \frac{1}{2}AB$.

Now $EB \parallel FC$ and $EB = FC$ $\Rightarrow$ EBCF is a parallelogram $\Rightarrow$ $EF \parallel BC$ and $EC \parallel BF$.

Also $AE \parallel DF$ and $AE = DF$ $\Rightarrow$ AEFD is a parallelogram $\Rightarrow$ $AF \parallel ED$ and $AF = ED$.

Let $P$ = intersection of $AF$ and $BD$, $Q$ = intersection of $EC$ and $BD$.

In $\triangle ABP$: $E$ is mid-point of $AB$ and $EQ \parallel AP$ (since $EC \parallel AF$, so $EQ \parallel AP$).
By mid-point theorem converse: $Q$ is mid-point of $BP$ ... (i)

In $\triangle DCQ$: $F$ is mid-point of $DC$ and $FP \parallel CQ$ (since $AF \parallel EC$, so $FP \parallel QC$).
By mid-point theorem converse: $P$ is mid-point of $DQ$ ... (ii)

From (i): $BQ = QP$
From (ii): $DP = PQ$

So $BQ = QP = PD$.

Hence AF and EC trisect the diagonal BD. $\blacksquare$

Given: $\triangle ABC$ is right-angled at C; M is the mid-point of hypotenuse AB; $MD \parallel BC$, where D is on AC.

(i) D is the mid-point of AC:

In $\triangle ABC$:

• M is the mid-point of AB
• $MD \parallel BC$ (given)

By the Converse of Mid-point Theorem, D is the mid-point of AC. $\checkmark$

(ii) $MD \perp AC$:

Since $MD \parallel BC$ and $\angle ACB = 90^\circ$ (given):
$$\angle MDC = \angle BCD = 90^\circ \quad \text{(corresponding angles, } MD \parallel BC\text{)}$$

Wait — $\angle ACB = 90^\circ$ means $BC \perp AC$. Since $MD \parallel BC$:
$$\angle MDA = \angle BCA = 90^\circ \quad \text{(corresponding angles)}$$

Therefore $MD \perp AC$. $\checkmark$

(iii) $CM = MA = \frac{1}{2}AB$:

$MA = \frac{1}{2}AB$ (since M is the mid-point of AB). $\checkmark$

Now we need to show $CM = MA$.

In $\triangle ADM$ and $\triangle CDM$:

• $AD = CD$ (D is mid-point of AC, proved in (i))
• $\angle ADM = \angle CDM = 90^\circ$ (proved in (ii))
• $DM = DM$ (common)

By SAS: $\triangle ADM \cong \triangle CDM$

$\therefore CM = AM$ (CPCT)

Hence:
$$CM = MA = \frac{1}{2}AB \blacksquare$$