*Q1. The angles of quadrilateral are in the ratio \(3:5:9:13\). Find the angles of the quadrilateral.*

**Solution:**

Let the common ratio between the angles be x.

We know that the ‘Sum of the interior angles of the quadrilateral’ = \(360^{\circ}\)

Now,

\(3x+5x+9x+13x=360^{\circ}\)

\(\Rightarrow 30x=360^{\circ}\)

\(\Rightarrow x=12^{\circ}\)

Therefore the Angles of the quadrilateral are:

\(3x=3\times 12=36^{\circ}\)

*Q2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
*

**Solution:**

**Given,**

PQ = RS

**To show,**

PQRS is a rectangle we have to prove that one of its interior angle is right angled.

**Proof,**

In \(\bigtriangleup PQR\;\; and \;\; \bigtriangleup QPS\),

\(QR=QP\) (Common side)

\(PR=PS\) (Opposite sides of a parallelogram are equal)

\(PR=QS\) (Given)

Therefore, \(\bigtriangleup PQR \cong \bigtriangleup QPS\)by SSS congruence condition.

∠P = ∠Q (by CPCT)

also,

∠P + ∠Q = 180° (Sum of the angles on the same side of the transversal)

⇒ 2∠P = 180°

⇒ ∠P = 90°

Thus PQRS is a rectangle.

*Q3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.*

**Solution:**

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.

**Given,**

\(OA=OC, OB=OD \;\; and \;\; \angle AOB=\angle BOC=\angle OCD=\angle ODA=90^{\circ}\)

**To show,**

ABCD is parallelogram and \(AB=BC=CD=DA\)

**Proof,**

In \(\bigtriangleup AOB \;\; and \;\;\bigtriangleup COB\),

\(OA=OC\) (Given)

\(\angle AOB=\angle COB\) (Opposite sides of a parallelogram are equal)

\(OB=BO\) (Common)

Therefore, \(\bigtriangleup AOB \cong bigtriangleup COB\) (by SAS congruence condition).

Thus, \(AB=BC\) (by CPCT-Corresponding parts of Congruent)

Similarly we can prove,

\(AB=BC=CD=DA\)

Opposites sides of a quadrilateral are equal hence \(ABCD\) is a parallelogram.

Thus, \(ABCD\) is rhombus as it is a parallelogram whose diagonals intersect at right angle.

*Q4. Show that the diagonals of a square are equal and bisect each other at right angles.*

**Solution:**

Let PQRS be a square and its diagonals PR and QS intersect each other at O.

**To show,**

\(PR=QS,PO=OR\;\; and \;\; \angle POQ=90^{\circ}\)

**Proof,**

In \(\bigtriangleup PQR\;\; and \;\;\bigtriangleup QRS\),

\(QR=QP\) (Common)

\(\angle PQR=\angle QPS=90^{\circ}\)

\(PR=PS\) (Given)

Therefore, \(\bigtriangleup PQR \cong \bigtriangleup QRS\) by SAS congruence condition.

Thus, \(PR=PS\) (by CPCT).

Therefore, diagonals are equal.

Now,

In \(\bigtriangleup POQ\;\; and \;\;\bigtriangleup ROS\),

\(\angle QPO=\angle SRO\) (Alternate interior angles)

\(\angle POQ=\angle ROS\) (Vertically opposite)

\(PQ=RS\) (Given)

Therefore, ΔAOB ≅ ΔCOD (by AAS congruence condition).

Thus, \(PO=RO\) by CPCT. (Diagonal bisect each other.)

Now,

In \(\bigtriangleup POQ\;\; and \;\;\bigtriangleup ROQ\),

\(OQ=QO\) (Given)

\(PO=RO\) (diagonals are bisected)

\(PQ=RQ\) (Sides of the square)

Therefore, \(\bigtriangleup POQ \cong \bigtriangleup ROQ\) by SSS congruence condition.

also, \(\angle POQ=\angle ROQ\)

\(\angle POQ+\angle ROQ=180^{\circ}\) (Linear pair)

Thus, \(\angle POQ=\angle ROQ=90^{\circ}\) (Diagonals bisect each other at right angles)

*Q5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
*

**Solution:**

**Given,**

Let \(ABCD\) be a quadrilateral in which diagonals \(AC\;\; and \;\;BD\) bisect each other at right angle at O.

**To prove,**

Quadrilateral \(ABCD\) is a square.

**Proof,**

In \(\bigtriangleup AOB\;\; and \;\;\bigtriangleup COD\),

\(AO=CO\) (Diagonals bisect each other)

\(\angle AOB=\angle COD\) (Vertically opposite)

\(OB=OD\) (Diagonals bisect each other)

Therefore, \(\bigtriangleup AOB \cong \bigtriangleup COD\)by SAS congruence condition.

Thus, \(AB=CD\) by CPCT. …………… (i)

also,

\(\angle OAB=\angle OCD\) (Alternate interior angles)

⇒ \(\Rightarrow AB\parallel CD\)

Now,

In \(\bigtriangleup AOD\;\; and \;\;\bigtriangleup COD\),

\(AO=CO\) (Diagonals bisect each other)

\(\angle AOD=\angle COD\) (Vertically opposite)

\(OD=OD\) (Common)

Therefore, \(\bigtriangleup AOD \cong \bigtriangleup COD\) (by SAS congruence condition).

Thus, \(AD=CD\) (by CPCT). ………………… (ii)

also,

\(AD=BC\;\; and \;\;AD=CD\)

\(\Rightarrow AD=BC=CD=AB\) ……………….(iii)

also, \(\angle ADC=\angle BCD\) (by CPCT).

and \(\angle ADC+\angle BCD=180^{\circ}\) (co-interior angles)

\(\Rightarrow 2\angle ADC=180^{\circ}\)

\(\Rightarrow \angle ADC=90^{\circ}\) ……………. (iv)

One of the interior angle is right angle.

Thus, from (i), (ii) , (iii) and (iv) the given quadrilateral ABCD is a square.

*Q6. Diagonal \(AC\) of a parallelogram \(ABCD\) bisects \(\angle A\). Show that
(i) it bisects \(\angle C\) also,
(ii) \(ABCD\) is a rhombus.*

**Solution:**

(i)In \(\bigtriangleup ADC\;\;and \;\;\bigtriangleup CBA\),

\(AD=CB\) (Opposite sides of a parallelogram)

\(DC=BA\) (Opposite sides of a parallelogram)

\(AC=CA\) (Common)

Therefore, \(\bigtriangleup ADC \cong \bigtriangleup CBA\) by SSS congruence condition.

Thus,

\(\angle ACD=\angle CAB\) (by CPCT)

and \(\angle CAB=\angle CAD\) (Given)

\(\Rightarrow \angle ACD=\angle BCA\)

Thus, \(AC\) bisects \(\angle C\) also.

(ii) \(\angle ACD=\angle CAD\) (Proved)

⇒ \(AD=CD\) (Opposite sides of equal angles of a triangle are equal)

Also, \(AB=BC=CD=DA\) (Opposite sides of a parallelogram)

Thus, \(ABCD\) is a rhombus.

*Q7. \(ABCD\) is a rhombus. Show that diagonal \(AC\) bisects \(\angle A\) as well as \(\angle C\) and diagonal \(BD\) bisects \(\angle B\) as well as \(\angle D\).*

**Solution:**

**Let **\(ABCD\) is a rhombus and \(AC\;\; and\;\; BD\) are its diagonals.

**Proof,**

\(AD=CD\) (Sides of a rhombus)

\(\angle DAC=\angle DCA\) (Angles opposite of equal sides of a triangle are equal.)

also, \(AB\parallel CD\)

\(\Rightarrow \angle DAC=\angle BCA\) (Alternate interior angles)

\(\Rightarrow \angle DCA=\angle BCA\)

Therefore, \(AC\;\; bisect\;\; \angle C\).

Similarly, we can prove that diagonal \(AC\;\; bisect\;\; \angle A\).

Also, by preceding above method we can prove that diagonal \(BD\;\; bisect\;\; \angle B\;\; as \;\; well \;\; as \angle D\).

*Q8. \(PQRS\) is a rectangle in which diagonal \(PR\) bisects \(\angle P\) as well as \(\angle R\). Show that:
(i) \(PQRS\) is a square
(ii) diagonal \(QS\) bisects \(\angle Q\) as well as \(\angle S\).*

** Solution:**

(i) \(\angle SPR=\angle SRP\) (\(PR\) bisects \(\angle P\) as well as \(\angle R\))

\(\Rightarrow PS=RS\) (Sides opposite to equal angles of a triangle are equal)

also, \(RS=PQ\) (Opposite sides of a rectangle)

Therefore, \(PQ=QR=RS=SP\)

Thus, \(PQRS\) is a square.

(ii) In \(\bigtriangleup QRS\),

\(QR=RS\)

\(\Rightarrow \angle RSQ=\angle RQS\) (Angles opposite to equal sides are equal)

also, \(\angle RSQ=\angle PQS\) (Alternate interior angles)

\(\Rightarrow \angle RQS=\angle PQS\)

Thus, \(QS\) bisects \(\angle Q\).

Now,

\(\angle RQS=\angle PSQ\)

\(\Rightarrow \angle RSQ=\angle PSQ\)

Thus, \(BD\) bisects \(\angle D\)

*Q9. In parallelogram \(ABCD\), two points \(P\;\; and \;\;Q\) are taken on diagonal \(BD\) such that \(DP=BQ\). Show that:*

(i) \(\bigtriangleup APD \cong \bigtriangleup CQB\)

(ii) \(AP=CQ\)

(iii) \(\bigtriangleup AQB \cong \bigtriangleup CPD\)

(iv) \(AQ=CP\)

(v) \(APCQ\) is a parallelogram

**Solution:**

(i) In \(\bigtriangleup APD \;\; and \;\; \bigtriangleup CQB\),

\(DP=BQ\) (Given)

\(\angle ADP=\angle CBQ\) (Alternate interior angles)

\(AD=BC\) (Opposite sides of a ||gm)

Thus, \(\bigtriangleup APD \cong \bigtriangleup CQB\) (by SAS congruence condition).

(ii) \(AP=CQ\) by CPCT as \(\bigtriangleup APD \cong \bigtriangleup CQB\).

(iii) In \(\bigtriangleup AQB\;\; and \;\; \bigtriangleup CPD\),

\(BQ=DP\) (Given)

\(\angle ABQ=\angle CDP\) (Alternate interior angles)

\(AB=BC=CD\) (Opposite sides of a parallelogram)

Thus, \(\bigtriangleup AQB \cong \bigtriangleup CPD\) (by SAS congruence condition).

(iv) \(AQ=CP\) by CPCT as \(\bigtriangleup AQB \cong \bigtriangleup CPD\).

(v) From (ii) and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus, \(APCQ\) is a parallelogram.

*10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that
(i) \(\bigtriangleup APB \cong \bigtriangleup CQD\)
(ii) AP = CQ*

**Solution:**

(i) In \(\bigtriangleup APB \;\; and \;\; \bigtriangleup CQD\),

\(\angle ABP=\angle CDQ\) (Alternate interior angles)

\(\angle APB=\angle CQD\) (equal to right angles as AP and CQ are perpendiculars)

AB = CD (ABCD is a parallelogram)

Thus, \(\bigtriangleup APB \cong \bigtriangleup CQD\) (by AAS congruence condition).

(ii) AP = CQ by CPCT as \(\bigtriangleup APB \cong \bigtriangleup CQD\).

*Q11. In \(\bigtriangleup ABC\;\; and \;\;\bigtriangleup DEF\), \(AB=DE,\; AB\parallel DE,\;BC=EF\; and \;BC\parallel EF\). Vertices A, B and C are joined to vertices D, E and F respectively.
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) \(AD\parallel CF\) and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) \(AC=DF\) *

*(vi) \(\bigtriangleup ABC \cong \bigtriangleup DEF\)*

**Solution:**

(i) \(AB=DE\;\; and \;\; AB\parallel DE\) (Given)

Thus, quadrilateral ABED is a parallelogram because two opposite sides of a quadrilateral are equal and parallel to each other.

(ii) Again \(BC=EF\;\; and \;\; BC\parallel EF\).

Thus, quadrilateral BEFC is a parallelogram.

(iii) Since ABED and BEFC are parallelograms.

\(\Rightarrow AD=BE\; and\; BE=CF\) (Opposite sides of a parallelogram are equal)

Thus, \(AD=CF\).

Also, \(AD\parallel BE\;\; and \;\; BE\parallel CF\) (Opposite sides of a parallelogram are parallel)

Thus, \(AD\parallel CF\).

(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.

(v) \(AC\parallel DF\;\; and \;\; AC=DF\) because ACFD is a parallelogram.

(vi) \(\bigtriangleup ABC\;\; and\;\;\bigtriangleup DEF\),

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

Thus, \(\bigtriangleup ABC\cong\bigtriangleup DEF\) (by SSS congruence condition).

*Q12. 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) \(\bigtriangleup ABC \cong \bigtriangleup BAD\)

(iv) diagonal AC = diagonal BD

*[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]*

**Solution:**

**Construction**: Draw a line through C parallel to DA intersecting AB produced at E.

**(i)** CE = AD (Opposite sides of a parallelogram)

AD = BC (Given)

Therefor, BC = CE

\(\Rightarrow \angle CBE=\angle CEB\)also,

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

\(\angle B+\angle CBE=180^{\circ}\) (Linear pair)

\(\Rightarrow \angle A=\angle B\)

**(ii)** \(\angle A+\angle D=\angle B+\angle C=180^{\circ}\) (Angles on the same side of transversal)

\(\Rightarrow \angle A+\angle D=\angle A+\angle C\) (as, \(\angle A=\angle B\))

**(iii)** In \(\bigtriangleup ABC\;\; and \;\; \bigtriangleup BAD\),

AB = AB (Common)

\(\angle DBA =\angle CBA\)

AD = BC (Given)

Thus, \(\bigtriangleup ABC \cong \bigtriangleup BAD\) (by SAS congruence condition).

**(iv) ** Diagonal AC = diagonal BD (by CPCT as \(\bigtriangleup ABC \cong \bigtriangleup BAD\).)

**Exercise 2:**

*Q1.**\(ABCD\) 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.

**Solution:**

(i) In \(\bigtriangleup DAC\),

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

Thus by mid point theorem, \(SR\parallel AC\) and \(SR=\frac{1}{2}AC\)

(ii) In \(\bigtriangleup BAC\),

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

Thus by mid point theorem, \(PQ\parallel AC\) and \(PQ=\frac{1}{2} AC\)

also, \(SR=\frac{1}{2} AC\)

Thus, PQ = SR

(iii) \(SR\parallel AC\)– from (i)

and, \(PQ\parallel AC\)– from (ii)

\(\Rightarrow SR \parallel PQ\)– from (i) and (ii)

also, PQ = SR

Thus, PQRS is a parallelogram.

*Q2. 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.*

**Solution:**

**Given,**

ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

**To Prove,**

PQRS is a rectangle.

**Construction,**

Join AC and BD

**Proof,**

In \(\bigtriangleup DRS \;\;and\;\;\bigtriangleup BPQ\),

DS = BQ (Halves of the opposite sides of the rhombus)

\(\angle SDR=\angle QBP\) (Opposite angles of the rhombus)

DR = BP (Halves of the opposite sides of the rhombus)

Thus, \(\bigtriangleup DRS \cong \bigtriangleup BPQ\) (by SAS congruence condition).

RS = PQ (by CPCT) ………… (i)

In \(\bigtriangleup QCR \;\; and \;\; \bigtriangleup SAP\),

RC = PA (Halves of the opposite sides of the rhombus)

\(\angle RCQ=\angle PAS\) (Opposite angles of the rhombus)

CQ = AS (Halves of the opposite sides of the rhombus)

Thus, \(\bigtriangleup QCR \cong \bigtriangleup SAP\) (by SAS congruence condition).

RQ = SP (by CPCT) ……(ii)

Now,

In \(\bigtriangleup CBD\),

R and Q are the mid points of CD and BC respectively.

\(\Rightarrow QR \parallel BD\)

also,

P and S are the mid points of AD and AB respectively.

\(\Rightarrow PS \parallel BD\)

\(\Rightarrow QR \parallel PS\)

Thus, PQRS is a parallelogram.

also, \(\angle PQR=90^{\circ}\)

Now,

In PQRS,

RS = PQ and RQ = SP from (i) and (ii)

\(\angle Q=90^{\circ}\)

Thus, PQRS is a rectangle.

*Q3. 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.
*

**Solution:**

**Given,**

ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

**Construction,**

AC and BD are joined.

**To Prove,**

PQRS is a rhombus.

**Proof,**

In \(\bigtriangleup ABC\)

P and Q are the mid-points of AB and BC respectively

Thus, \(PQ\parallel AC\) and \(PQ= \frac{1}{2}AC\) (Mid point theorem) ……….(i)

In \(\bigtriangleup ADC\),

\(SR\parallel AC\) and \(SR=\frac{1}{2}AC\) (Mid point theorem) ……(ii)

So, \(PQ\parallel SR\) and \(PQ= SR\)

As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.

\(PS\parallel QR\) and \(PS= QR\) (Opposite sides of parallelogram) ….(iii)

Now,

In \(\bigtriangleup BCD\),

Q and R are mid points of side BC and CD respectively.

Thus, \(QR\parallel BD\;\; and\;\; QR=\frac{1}{2}BD\) (Mid point theorem) ……(iv)

AC = BD (Diagonals of a rectangle are equal) ……… (v)

From equations (i), (ii), (iii), (iv) and (v),

PQ = QR = SR = PS

So, PQRS is a rhombus.

*Q4. 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.*

** **

**Solution:**

**Given,**

ABCD is a trapezium in which \(AB\parallel DC\), BD is a diagonal and E is the mid-point of AD.

**To prove,**

F is the mid-point of BC.

**Proof,**

BD intersected EF at G.

In \(\bigtriangleup BAD\),

E is the mid point of AD and also \(EG\parallel AB\).

Thus, G is the mid point of BD (Converse of mid point theorem)

Now,

In \(\bigtriangleup BDC\),

G is the mid point of BD and also \(GF\parallel AB\parallel DC\).

Thus, F is the mid point of BC (Converse of mid point theorem)

*Q5. 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.*

* *

**Solution:**

**Given**

ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.

**To show,**

AF and EC trisect the diagonal BD.

**Proof,**

ABCD is a parallelogram

Therefore, \(AB\parallel CD\)

also, \(AE\parallel FC\)

Now,

\(AB=CD\) (Opposite sides of parallelogram ABCD)

\(\Rightarrow \frac{1}{2}AB=\frac{1}{2}CD\)

\(\Rightarrow AE=FC\) (E and F are midpoints of side AB and CD)

AECF is a parallelogram (AE and CF are parallel and equal to each other)

\(AF\parallel EC\) (Opposite sides of a parallelogram)

Now,

In \(\bigtriangleup DQC\),

F is mid point of side DC and \(FP\parallel CQ\) (as \(AF\parallel EC\)).

P is the mid-point of DQ (Converse of mid-point theorem)

\(\Rightarrow DP=PQ\) ………(i)

Similarly,

In \(\bigtriangleup APB\),

E is mid point of side AB and \(EQ\parallel AP\) (as \(AF\parallel EC\)).

Q is the mid-point of PB (Converse of mid-point theorem)

\(\Rightarrow PQ=QB\) …….(ii)

From equations (i) and (i),

DP = PQ = BQ

Hence, the line segments AF and EC trisect the diagonal BD.

*Q6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
*

**Solution:**

Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively.

Now,

In \(\bigtriangleup ACD\),

R and S are the mid points of CD and DA respectively.

Thus, \(SR\parallel AC\).

Similarly we can show that,

\(PQ\parallel AC\)

\(PS\parallel BD\)

\(QR\parallel BD\)

Thus, PQRS is parallelogram.

PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

*Q7. 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) \(CM=MA=\frac{1}{2}AB\)*

**Solution:**

(i) In \(\bigtriangleup ACB\),

M is the mid point of AB and \(MD\parallel BC\)

Thus, D is the mid point of AC (Converse of mid point theorem)

(ii) \(\angle ACB=\angle ADM\) (Corresponding angles)

also, \(\angle ACB=90^{\circ}\)

Thus, \(\angle ADM=90^{\circ}\)and \(MD\perp AC\)

(iii) In \(\bigtriangleup AMD\;\; and\;\; \bigtriangleup CMD\),

AD = CD (D is the midpoint of side AC)

\(\angle ADM=\angle CDM\) (Each 90°)

DM = DM (common)

Thus, \(\bigtriangleup AMD \cong \bigtriangleup CMD\) (by SAS congruence condition).

AM = CM (by CPCT)

also, \(AM=\frac{1}{2}AB\) (as M is mid point of AB)

Hence, \(CM=AM=\frac{1}{2}AB\).

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