Quadrilaterals

Given: ABCD is a parallelogram in which diagonal AC = diagonal BD.

To prove: ABCD is a rectangle.

Proof:

In $\triangle ABC$ and $\triangle DCB$:
- $AB = DC$ (opposite sides of a parallelogram)
- $BC = BC$ (common)
- $AC = DB$ (given, diagonals are equal)

By SSS congruence rule:
$$\triangle ABC \cong \triangle DCB$$

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

Since $AB \parallel DC$ and $BC$ is a transversal:
$$\angle ABC + \angle DCB = 180^\circ \quad \text{(co-interior angles)}$$

$$\Rightarrow 2\angle ABC = 180^\circ$$

$$\Rightarrow \angle ABC = 90^\circ$$

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

Given: ABCD is a square, i.e., $AB = BC = CD = DA$ and all angles are $90^\circ$.

To prove: (a) $AC = BD$, (b) diagonals bisect each other, (c) they bisect at right angles.

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(a) AC = BD:

In $\triangle ABC$ and $\triangle BAD$:
- $AB = BA$ (common)
- $BC = AD$ (sides of a square)
- $\angle ABC = \angle BAD = 90^\circ$

By SAS: $\triangle ABC \cong \triangle BAD$

$\therefore AC = BD$ (CPCT) $\checkmark$

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(b) Diagonals bisect each other:

Let diagonals AC and BD intersect at O.

In $\triangle AOB$ and $\triangle COD$:
- $AB = CD$ (sides of a square)
- $\angle OAB = \angle OCD$ (alternate interior angles, $AB \parallel CD$)
- $\angle OBA = \angle ODC$ (alternate interior angles)

By AAS: $\triangle AOB \cong \triangle COD$

$\therefore OA = OC$ and $OB = OD$ (CPCT)

Hence diagonals bisect each other. $\checkmark$

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(c) Diagonals bisect at right angles:

In $\triangle AOB$ and $\triangle AOD$:
- $OB = OD$ (proved above)
- $AB = AD$ (sides of a square)
- $AO = AO$ (common)

By SSS: $\triangle AOB \cong \triangle AOD$

$\therefore \angle AOB = \angle AOD$ (CPCT)

But $\angle AOB + \angle AOD = 180^\circ$ (linear pair)

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

Hence diagonals bisect each other at right angles. $\blacksquare$

Given: ABCD is a parallelogram in which diagonal AC bisects $\angle A$, i.e., $\angle DAC = \angle BAC$.

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(i) AC bisects $\angle C$:

Since $AB \parallel DC$ and $AC$ is a transversal:
$$\angle BAC = \angle DCA \quad \text{(alternate interior angles)} \quad \ldots(1)$$

Since $AD \parallel BC$ and $AC$ is a transversal:
$$\angle DAC = \angle BCA \quad \text{(alternate interior angles)} \quad \ldots(2)$$

But $\angle DAC = \angle BAC$ (given) $\ldots(3)$

From (1), (2) and (3):
$$\angle DCA = \angle BCA$$

Therefore, AC bisects $\angle C$ also. $\checkmark$

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(ii) ABCD is a rhombus:

In $\triangle ABC$:
$$\angle BAC = \angle BCA \quad \text{(from (1) and (3))}$$

$\Rightarrow BC = AB$ (sides opposite equal angles are equal)

Since ABCD is a parallelogram, $AB = CD$ and $BC = AD$.

Therefore $AB = BC = CD = DA$.

Hence ABCD is a rhombus. $\blacksquare$

Given: ABCD is a rectangle; AC bisects $\angle A$ and $\angle C$.

So $\angle DAC = \angle BAC = 45^\circ$ and $\angle DCA = \angle BCA = 45^\circ$ (since each angle of a rectangle is $90^\circ$).

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(i) ABCD is a square:

In $\triangle ABC$:
$$\angle BAC = 45^\circ \text{ and } \angle BCA = 45^\circ$$
$$\Rightarrow \angle BAC = \angle BCA$$
$$\Rightarrow BC = AB \quad \text{(sides opposite equal angles)}$$

Since ABCD is a rectangle, $AB = CD$ and $BC = AD$.

Therefore $AB = BC = CD = DA$.

A rectangle with all sides equal is a square. $\checkmark$

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(ii) BD bisects $\angle B$ and $\angle D$:

Since ABCD is a square, $AB = BC = CD = DA$.

In $\triangle ABD$:
$$AB = AD \Rightarrow \angle ABD = \angle ADB$$

Also $\angle ABD + \angle ADB = 90^\circ$ (since $\angle A = 90^\circ$)
$$\Rightarrow \angle ABD = \angle ADB = 45^\circ = \frac{1}{2} \times 90^\circ$$

So BD bisects $\angle B$.

Similarly in $\triangle BCD$: $BC = CD \Rightarrow \angle CBD = \angle CDB = 45^\circ$

So BD bisects $\angle D$ also. $\blacksquare$

Given: ABCD is a parallelogram; P and Q are on diagonal BD such that $DP = BQ$.

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(i) $\triangle APD \cong \triangle CQB$:

In $\triangle APD$ and $\triangle CQB$:
- $AD = CB$ (opposite sides of parallelogram)
- $DP = BQ$ (given)
- $\angle ADP = \angle CBQ$ (alternate interior angles, since $AD \parallel BC$ and $BD$ is transversal)

By SAS: $\triangle APD \cong \triangle CQB$ $\checkmark$

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(ii) $AP = CQ$:

From (i), $AP = CQ$ (CPCT) $\checkmark$

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(iii) $\triangle AQB \cong \triangle CPD$:

In $\triangle AQB$ and $\triangle CPD$:
- $AB = CD$ (opposite sides of parallelogram)
- $BQ = DP$ (given)
- $\angle ABQ = \angle CDP$ (alternate interior angles, since $AB \parallel CD$ and $BD$ is transversal)

By SAS: $\triangle AQB \cong \triangle CPD$ $\checkmark$

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(iv) $AQ = CP$:

From (iii), $AQ = CP$ (CPCT) $\checkmark$

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(v) APCQ is a parallelogram:

From (ii): $AP = CQ$
From (iv): $AQ = CP$

Since both pairs of opposite sides are equal, APCQ is a parallelogram. $\blacksquare$

Given: ABCD is a parallelogram; $AP \perp BD$ and $CQ \perp BD$.

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(i) $\triangle APB \cong \triangle CQD$:

In $\triangle APB$ and $\triangle CQD$:
- $\angle APB = \angle CQD = 90^\circ$ (given, AP and CQ are perpendiculars)
- $AB = CD$ (opposite sides of parallelogram)
- $\angle ABP = \angle CDQ$ (alternate interior angles, since $AB \parallel CD$ and $BD$ is transversal)

By AAS: $\triangle APB \cong \triangle CQD$ $\checkmark$

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(ii) $AP = CQ$:

From (i), $AP = CQ$ (CPCT) $\blacksquare$

Given: ABCD is a trapezium with $AB \parallel CD$ and $AD = BC$.

Construction: Extend AB to E and draw $CE \parallel DA$, meeting AB produced at E.

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(i) $\angle A = \angle B$:

Since $AD \parallel CE$ and $AE \parallel DC$ (by construction and given), ADCE is a parallelogram.

$\therefore AD = CE$ (opposite sides of parallelogram)

But $AD = BC$ (given), so $BC = CE$.

In $\triangle BCE$: $BC = CE \Rightarrow \angle CBE = \angle CEB$ (base angles of isosceles triangle)

$\angle CBE = \angle ABC$ (same angle)

Now, $\angle DAB + \angle CBE = 180^\circ$ ... wait, let us use co-interior angles.

Since $AD \parallel CE$:
$$\angle DAE + \angle CEA = 180^\circ \quad \text{(co-interior angles on same side of AE)}$$

In $\triangle BCE$, $BC = CE$, so $\angle CBE = \angle CEB$.

$$\angle ABC = 180^\circ - \angle CBE = 180^\circ - \angle CEB$$

Also $\angle DAB + \angle CEB = 180^\circ$ (since $\angle DAE = \angle DAB$ and $\angle CEA = \angle CEB$, co-interior angles with $AD \parallel CE$)

$$\Rightarrow \angle DAB = 180^\circ - \angle CEB = \angle ABC$$

$$\therefore \angle A = \angle B \checkmark$$

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(ii) $\angle C = \angle D$:

Since $AB \parallel CD$:
$$\angle A + \angle D = 180^\circ \quad \text{and} \quad \angle B + \angle C = 180^\circ$$

Since $\angle A = \angle B$:
$$\angle D = 180^\circ - \angle A = 180^\circ - \angle B = \angle C$$

$$\therefore \angle C = \angle D \checkmark$$

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(iii) $\triangle ABC \cong \triangle BAD$:

In $\triangle ABC$ and $\triangle BAD$:
- $AB = BA$ (common)
- $BC = AD$ (given)
- $\angle ABC = \angle BAD$ (proved in (i))

By SAS: $\triangle ABC \cong \triangle BAD$ $\checkmark$

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(iv) diagonal AC = diagonal BD:

From (iii): $AC = BD$ (CPCT) $\blacksquare$

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

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(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:
$$SR \parallel AC \quad \text{and} \quad SR = \frac{1}{2}AC \checkmark$$

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

$$\therefore PQ = SR \checkmark$$

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(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.

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(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. $\checkmark$

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(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 $MD \perp AC$. $\checkmark$

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