Maths 4 Marks Questions

1. In $\triangle\text{APD}$ and $\triangle\text{CQB}$

$\angle\text{ADP}=\angle\text{CBQ}$ (Alternate interior angles for BC || AD)

AD = CB (Opposite sides of parallelogram ABCD)

DP = BQ (Given)

$\Rightarrow \triangle\text{APQ}\cong\triangle\text{CQB}$ (Using SAS congruence rule)

2. As we had observed that $\Rightarrow \triangle\text{APQ}\cong\triangle\text{CQB}$

 AP = CQ (CPCT)

3. In $\triangle\text{AQB}$ and $\triangle\text{CQB}$

$\angle\text{ABQ}=\angle\text{CPD}$ (Alternate interior angles for AB || CD)

AB = CD (Opposite sides of parallelogram ABCD)

BQ = DP (Given)

$\Rightarrow \triangle\text{APQ}\cong\triangle\text{CQB}$ (Using SAS congruence rule)

4. As we had observed that $\Rightarrow \triangle\text{APQ}\cong\triangle\text{CQB}$

AQ = CP (CPCT)

5. From the result obtained in (ii) and (iv),

AQ = CP and

AP = CQ

Since opposite sides in quadrilateral APCQ are equal to each other, APCQ is a parallelogram.

1. It is given that AB = DE and AB || DE.

If two opposite sides of a quadrilateral are equal and parallel to each other, then it will be a parallelogram.

Therefore, quadrilateral ABED is a parallelogram.

2. Again, BC = EF and BC || EF

Therefore, quadrilateral BCEF is a parallelogram.

3. As we had observed that ABED and BEFC are parallelograms, therefore

AD = BE and AD || BE

(Opposite sides of a parallelogram are equal and parallel)

And, BE = CF and BE || CF

(Opposite sides of a parallelogram are equal and parallel)

 AD = CF and AD || CF

4. As we had observed that one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and parallel to each other, therefore, it is a parallelogram.

5. As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal and parallel to each other.

 AC || DF and AC = DF

6. $\therefore \triangle\text{ABC}\cong\triangle\text{DEF}$

AB = DE (Given)

BC = EF (Given)

AC = DF (ACFD is a parallelogram)

$\therefore \triangle\text{ABC}\cong\triangle\text{DEF}$ (By SSS congruence rule).

1. CE = AD (Opposite sides of a parallelogram)

AD = BC (Given)

Therefor, BC = CE

also,

$\angle\text{A}+\angle\text{CBE}=180^{0}$ (Angles on the same side of transversal and ∠CBE = ∠CEB)

$\angle\text{B}+\angle\text{CBE}=180^{0}$ (Linear pair)

$\angle\text{A}=\angle\text{B}$

2. $\angle\text{A}+\angle\text{D}=\angle\text{B}+\angle\text{C}=180^{0}$ (Angles on the same side of transversal)

$\angle\text{A}+\angle\text{D}=\angle\text{B}+\angle\text{C}$

$\angle\text{D}=\angle\text{C}$

3. In $\angle\text{ABC}$ and $\angle\text{BAD}$

AB = AB (Common)

$\angle\text{DBA}=\angle\text{CBA}$

AD = BC (Given)

4. Thus, by SAS $\angle\text{DBA}=\angle\text{CBA}$ Congruence condition.

1. In $\triangle\text{ADC}$ and R are the mid-points of sides AD and CD respectively.

In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it.

SR || AC and SR $=\frac{1}{2}\text{AC}\ ...(\text{i})$

2. In $\triangle\text{ABC}$ and Q are mid - points of sides AB and BC respectively. Therefore, by using mid - point theorem,

PQ || AC and PQ $=\frac{1}{2}\text{AC}\ ...(\text{ii})$

Using equations (i) and (ii), we obtain

PQ || SR and PQ = SR ...(iii)

⇒ PQ = SR

3. From equation (3), we obtained

PQ || SR and PQ = SR

Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal.

Hence, PQRS is a parallelogram.

1. In $\triangle\text{ABC}$ and $\triangle\text{ADC},$ we have:

$\text{AB = AD}$ (Given)

$\text{BC = DC}$ (Given)

$\text{AC}$ is common.

i.e., $\triangle\text{ABC}\cong\triangle\text{ADC}$ (SSS congruence rule)

$\therefore\angle\text{BAC}=\angle\text{DAC}$ and $\angle\text{BCA}=\angle\text{DCA}$ (By C.P.C.T.)

Thus, AC bisects $\angle\text{A}$ and $\angle\text{C}.$

2. Now, in $\triangle\text{ABE}$ and $\triangle\text{ADE},$ we have:

$\text{AB = AD}$ (Given)

$\angle\text{BAE}=\angle\text{DAE}$ (Proven above)

$\text{AE}$ is common.

$\therefore\triangle\text{ABE}\cong\triangle\text{ADE}$ (SAS congruence rule)

$\Rightarrow\text{BE = DE}$ (By C.P.C.T.)

3. $\triangle\text{ABC}\cong\triangle\text{ADC}$ (Proven above)

$\therefore\angle\text{ABC}=\angle\text{ADC}$ (By C.P.C.T.)

Given in a parallelogram ABCD, P is a mid-point of BC such that $\angle\text{BAP}=\angle\text{DAP}.$

To prove AD = 2CD

Proof Since, ABCD is a parallelogram.

So, AD || BC and AB is transversal, then

$\angle\text{A}+\angle\text{B}=180^\circ$ (sum of cointerior angles is 180°)

$\angle\text{B}=180^\circ-\angle\text{A}\ ...(\text{i})$

in $\Delta\text{ABP}$ $\angle\text{PAB}+2\angle\text{B}+\angle\text{BPA}=180^\circ$[by angle sum property of a triangle)

$\Rightarrow\ \frac{1}{2}\angle\text{A}+180^\circ-\angle\text{A}+\angle\text{BPA}=180^\circ$ [from Eq. (i)]

$\Rightarrow\ \angle\text{BPA}=\frac{\angle\text{A}}{2}=0$

$\Rightarrow\ \angle\text{BPA}=\frac{\angle\text{A}}{2}\ ...\text{(ii)}$

$\Rightarrow\angle\text{BPA}=\angle\text{BAP}$

$\Rightarrow\ \text{AB}=\text{BP}$ [opposite sides of equal angles are equal]

On multiplying both sides by 2, we get

$2\text{AB}=2\text{BP}$

$\Rightarrow\ 2\text{AB}=\text{BP}$ (since Pis the mid-point of BC)

$\Rightarrow\ 2\text{CD}=\text{AD}$

(since, ABCD is a parallelogram, then AB = CD and BC = AD)

AB = AC and CD ∥ BA and AP is the bisector of exterior $\angle\text{CAD}$ of $\triangle\text{ABC}$

To prove:

1.  $\angle\text{PAC}=\angle\text{BCA}$

2.  ABCP is a parallelogram.

Proof:

1.  We have,

AB = AC

$\Rightarrow\angle\text{CAD}=\angle\text{ABC}$ [Opposite angles of equal sides of triangle are equal]

Now, $\angle\text{CAD}=\angle\text{ABC}+\angle\text{ACB}$

$\Rightarrow\angle\text{PAC}+\angle\text{PAD}=\angle\text{ACB}$ $[\therefore\angle\text{PAC}=\angle\text{PAD}]$ 

$\Rightarrow2\angle\text{PAC}=2\angle\text{ACB}$

$\Rightarrow\angle\text{PAC}=\angle\text{ACB}$

2. Now,

$\angle\text{PAC}=\angle\text{BCA}$

$\Rightarrow\angle\text{AP}||\text{BC}$ and $\text{CP}||\text{BA}$ [Given]

Therefore, ABCP is a parallelogram.

1. In $\triangle\text{APB}$ and $\triangle\text{CQD}$

$\angle\text{ABP}=\angle\text{CDQ}$ (Alternate interior angles)

$\angle\text{ABP}=\angle\text{CDQ}$ (equal to right angles as AP and CQ are perpendiculars)

AB = CD (ABCD is a parallelogram)

Thus, $\angle\text{ABP}=\angle\text{CDQ}$ by AAS congruence condition

2. AP = CQ by CPCT as $\angle\text{ABP}=\angle\text{CDQ}.$