# NCERT Solutions For Class 9 Maths Chapter 8

## NCERT Solutions Class 9 Maths Quadrilaterals

NCERT solutions for class 9 Maths chapter 8 Quadrilaterals include solutions for all the questions present in class 9 NCERT maths textbooks. This is the best study material for students preparing for their finals. NCERT solutions for class 9 Maths chapter 8 are available in a free PDF format to help students in practising Quadrilaterals on a regular basis. All the solutions present in this study materials are solved by the subject experts and are explained in a lucid manner along with the more examples, formulas, and other important questions. Students can download these NCERT solutions for class 9 Maths chapter 8 pdf files and prepare for their exams.

### NCERT Solutions For Class 8 Maths Chapter 8 Exercises

Q1. The angles of quadrilateral are in the ratio 3:5:9:13$3:5:9:13$. 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$360^{\circ}$
Now,
3x+5x+9x+13x=360$3x+5x+9x+13x=360^{\circ}$
30x=360$\Rightarrow 30x=360^{\circ}$
x=12$\Rightarrow x=12^{\circ}$
Therefore the Angles of the quadrilateral are:
3x=3×12=36$3x=3\times 12=36^{\circ}$

5x=5×12=60$5x=5\times 12=60^{\circ}$ 9x=9×12=108$9x=9\times 12=108^{\circ}$ 13x=13×12=156$13x=13\times 12=156^{\circ}$

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 PQRandQPS$\bigtriangleup PQR\;\; and \;\; \bigtriangleup QPS$,
QR=QP$QR=QP$ (Common side)
PR=PS$PR=PS$ (Opposite sides of a parallelogram are equal)
PR=QS$PR=QS$ (Given)
Therefore, PQRQPS$\bigtriangleup PQR \cong \bigtriangleup QPS$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=ODandAOB=BOC=OCD=ODA=90$OA=OC, OB=OD \;\; and \;\; \angle AOB=\angle BOC=\angle OCD=\angle ODA=90^{\circ}$

To show,
ABCD is parallelogram and AB=BC=CD=DA$AB=BC=CD=DA$
Proof,
In AOBandCOB$\bigtriangleup AOB \;\; and \;\;\bigtriangleup COB$,
OA=OC$OA=OC$ (Given)
AOB=COB$\angle AOB=\angle COB$ (Opposite sides of a parallelogram are equal)
OB=BO$OB=BO$ (Common)
Therefore, AOBbigtriangleupCOB$\bigtriangleup AOB \cong bigtriangleup COB$ (by SAS congruence condition).
Thus, AB=BC$AB=BC$ (by CPCT-Corresponding parts of Congruent)
Similarly we can prove,
AB=BC=CD=DA$AB=BC=CD=DA$
Opposites sides of a quadrilateral are equal hence ABCD$ABCD$ is a parallelogram.
Thus, ABCD$ABCD$ 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=ORandPOQ=90$PR=QS,PO=OR\;\; and \;\; \angle POQ=90^{\circ}$

Proof,
In PQRandQRS$\bigtriangleup PQR\;\; and \;\;\bigtriangleup QRS$,
QR=QP$QR=QP$ (Common)
PQR=QPS=90$\angle PQR=\angle QPS=90^{\circ}$
PR=PS$PR=PS$ (Given)
Therefore, PQRQRS$\bigtriangleup PQR \cong \bigtriangleup QRS$ by SAS congruence condition.
Thus, PR=PS$PR=PS$ (by CPCT).

Therefore, diagonals are equal.
Now,
In POQandROS$\bigtriangleup POQ\;\; and \;\;\bigtriangleup ROS$,
QPO=SRO$\angle QPO=\angle SRO$ (Alternate interior angles)
POQ=ROS$\angle POQ=\angle ROS$ (Vertically opposite)
PQ=RS$PQ=RS$ (Given)
Therefore, ΔAOB ≅ ΔCOD (by AAS congruence condition).
Thus, PO=RO$PO=RO$ by CPCT. (Diagonal bisect each other.)
Now,
In POQandROQ$\bigtriangleup POQ\;\; and \;\;\bigtriangleup ROQ$,
OQ=QO$OQ=QO$ (Given)
PO=RO$PO=RO$ (diagonals are bisected)
PQ=RQ$PQ=RQ$ (Sides of the square)
Therefore, POQROQ$\bigtriangleup POQ \cong \bigtriangleup ROQ$ by SSS congruence condition.
also, POQ=ROQ$\angle POQ=\angle ROQ$

POQ+ROQ=180$\angle POQ+\angle ROQ=180^{\circ}$ (Linear pair)
Thus, POQ=ROQ=90$\angle POQ=\angle ROQ=90^{\circ}$ (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$ABCD$ be a quadrilateral in which diagonals ACandBD$AC\;\; and \;\;BD$ bisect each other at right angle at O.
To prove,
Quadrilateral ABCD$ABCD$ is a square.
Proof,
In AOBandCOD$\bigtriangleup AOB\;\; and \;\;\bigtriangleup COD$,
AO=CO$AO=CO$ (Diagonals bisect each other)
AOB=COD$\angle AOB=\angle COD$ (Vertically opposite)
OB=OD$OB=OD$ (Diagonals bisect each other)
Therefore, AOBCOD$\bigtriangleup AOB \cong \bigtriangleup COD$by SAS congruence condition.
Thus, AB=CD$AB=CD$ by CPCT. …………… (i)
also,
OAB=OCD$\angle OAB=\angle OCD$ (Alternate interior angles)
ABCD$\Rightarrow AB\parallel CD$

Now,
In AODandCOD$\bigtriangleup AOD\;\; and \;\;\bigtriangleup COD$,
AO=CO$AO=CO$ (Diagonals bisect each other)
AOD=COD$\angle AOD=\angle COD$ (Vertically opposite)
OD=OD$OD=OD$ (Common)
Therefore, AODCOD$\bigtriangleup AOD \cong \bigtriangleup COD$ (by SAS congruence condition).
Thus, AD=CD$AD=CD$ (by CPCT).  ………………… (ii)
also,
AD=BCandAD=CD$AD=BC\;\; and \;\;AD=CD$
AD=BC=CD=AB$\Rightarrow AD=BC=CD=AB$  ……………….(iii)
also,  ADC=BCD$\angle ADC=\angle BCD$ (by CPCT).
and ADC+BCD=180$\angle ADC+\angle BCD=180^{\circ}$ (co-interior angles)
2ADC=180$\Rightarrow 2\angle ADC=180^{\circ}$
ADC=90$\Rightarrow \angle ADC=90^{\circ}$   ……………. (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$AC$ of a parallelogram ABCD$ABCD$ bisects A$\angle A$. Show that
(i) it bisects C$\angle C$ also,
(ii) ABCD$ABCD$ is a rhombus.

Solution:

(i)In ADCandCBA$\bigtriangleup ADC\;\;and \;\;\bigtriangleup CBA$,
AD=CB$AD=CB$ (Opposite sides of a parallelogram)
DC=BA$DC=BA$ (Opposite sides of a parallelogram)
AC=CA$AC=CA$ (Common)
Therefore, ADCCBA$\bigtriangleup ADC \cong \bigtriangleup CBA$ by SSS congruence condition.
Thus,
ACD=CAB$\angle ACD=\angle CAB$ (by CPCT)
and CAB=CAD$\angle CAB=\angle CAD$ (Given)
ACD=BCA$\Rightarrow \angle ACD=\angle BCA$
Thus, AC$AC$ bisects C$\angle C$ also.

(ii) ACD=CAD$\angle ACD=\angle CAD$ (Proved)
AD=CD$AD=CD$ (Opposite sides of equal angles of a triangle are equal)
Also, AB=BC=CD=DA$AB=BC=CD=DA$ (Opposite sides of a parallelogram)
Thus, ABCD$ABCD$ is a rhombus.

Q7. ABCD$ABCD$ is a rhombus. Show that diagonal AC$AC$ bisects A$\angle A$ as well as C$\angle C$ and diagonal BD$BD$ bisects B$\angle B$ as well as D$\angle D$.
Solution:

Let ABCD$ABCD$ is a rhombus and ACandBD$AC\;\; and\;\; BD$ are its diagonals.
Proof,
AD=CD$AD=CD$ (Sides of a rhombus)
DAC=DCA$\angle DAC=\angle DCA$ (Angles opposite of equal sides of a triangle are equal.)
also, ABCD$AB\parallel CD$
DAC=BCA$\Rightarrow \angle DAC=\angle BCA$ (Alternate interior angles)
DCA=BCA$\Rightarrow \angle DCA=\angle BCA$
Therefore, ACbisectC$AC\;\; bisect\;\; \angle C$.
Similarly, we can prove that diagonal ACbisectA$AC\;\; bisect\;\; \angle A$.

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

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

Solution:

(i) SPR=SRP$\angle SPR=\angle SRP$ (PR$PR$ bisects P$\angle P$ as well as R$\angle R$)
PS=RS$\Rightarrow PS=RS$ (Sides opposite to equal angles of a triangle are equal)
also, RS=PQ$RS=PQ$ (Opposite sides of a rectangle)
Therefore, PQ=QR=RS=SP$PQ=QR=RS=SP$
Thus, PQRS$PQRS$  is a square.

(ii) In QRS$\bigtriangleup QRS$,
QR=RS$QR=RS$
RSQ=RQS$\Rightarrow \angle RSQ=\angle RQS$ (Angles opposite to equal sides are equal)
also, RSQ=PQS$\angle RSQ=\angle PQS$ (Alternate interior angles)
RQS=PQS$\Rightarrow \angle RQS=\angle PQS$
Thus, QS$QS$ bisects Q$\angle Q$.
Now,
RQS=PSQ$\angle RQS=\angle PSQ$
RSQ=PSQ$\Rightarrow \angle RSQ=\angle PSQ$
Thus, BD$BD$ bisects D$\angle D$

Q9. In parallelogram ABCD$ABCD$, two points PandQ$P\;\; and \;\;Q$ are taken on diagonal BD$BD$ such that DP=BQ$DP=BQ$. Show that:
(i) APDCQB$\bigtriangleup APD \cong \bigtriangleup CQB$
(ii) AP=CQ$AP=CQ$
(iii) AQBCPD$\bigtriangleup AQB \cong \bigtriangleup CPD$
(iv) AQ=CP$AQ=CP$
(v) APCQ$APCQ$ is a parallelogram

Solution:

(i) In APDandCQB$\bigtriangleup APD \;\; and \;\; \bigtriangleup CQB$,
DP=BQ$DP=BQ$ (Given)
ADP=CBQ$\angle ADP=\angle CBQ$ (Alternate interior angles)
AD=BC$AD=BC$ (Opposite sides of a ||gm)
Thus, APDCQB$\bigtriangleup APD \cong \bigtriangleup CQB$ (by SAS congruence condition).

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

(iii) In AQBandCPD$\bigtriangleup AQB\;\; and \;\; \bigtriangleup CPD$,
BQ=DP$BQ=DP$ (Given)
ABQ=CDP$\angle ABQ=\angle CDP$ (Alternate interior angles)
AB=BC=CD$AB=BC=CD$ (Opposite sides of a parallelogram)
Thus, AQBCPD$\bigtriangleup AQB \cong \bigtriangleup CPD$ (by SAS congruence condition).

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

(v) From (ii)  and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus, APCQ$APCQ$ 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) APBCQD$\bigtriangleup APB \cong \bigtriangleup CQD$
(ii) AP = CQ

Solution:

(i) In APBandCQD$\bigtriangleup APB \;\; and \;\; \bigtriangleup CQD$,
ABP=CDQ$\angle ABP=\angle CDQ$ (Alternate interior angles)
APB=CQD$\angle APB=\angle CQD$ (equal to right angles as AP and CQ are perpendiculars)
AB = CD (ABCD is a parallelogram)
Thus, APBCQD$\bigtriangleup APB \cong \bigtriangleup CQD$ (by AAS congruence condition).

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

Q11. In ABCandDEF$\bigtriangleup ABC\;\; and \;\;\bigtriangleup DEF$, AB=DE,ABDE,BC=EFandBCEF$AB=DE,\; AB\parallel DE,\;BC=EF\; and \;BC\parallel EF$. 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) ADCF$AD\parallel CF$ and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC=DF$AC=DF$

(vi) ABCDEF$\bigtriangleup ABC \cong \bigtriangleup DEF$
Solution:

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

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

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

Thus, quadrilateral BEFC is a parallelogram.

(iii)  Since ABED and BEFC are parallelograms.
AD=BEandBE=CF$\Rightarrow AD=BE\; and\; BE=CF$ (Opposite sides of a parallelogram are equal)

Thus, AD=CF$AD=CF$.

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

Thus, ADCF$AD\parallel CF$.

(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.
(v) ACDFandAC=DF$AC\parallel DF\;\; and \;\; AC=DF$ because ACFD is a parallelogram.

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

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

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

Q12. ABCD is a trapezium in which ABCD$AB\parallel CD$ and AD=BC$AD=BC$ .Show that
(i) A=B$\angle A=\angle B$

(ii) C=D$\angle C=\angle D$
(iii) ABCBAD$\bigtriangleup ABC \cong \bigtriangleup BAD$
(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)
Therefor, BC = CE
CBE=CEB$\Rightarrow \angle CBE=\angle CEB$also,
A+CBE=180$\angle A+\angle CBE=180^{\circ}$ (Angles on the same side of transversal and CBE=CEB$\angle CBE=\angle CEB$)
B+CBE=180$\angle B+\angle CBE=180^{\circ}$ (Linear pair)
A=B$\Rightarrow \angle A=\angle B$

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

D=C$\Rightarrow \angle D=\angle C$

(iii) In ABCandBAD$\bigtriangleup ABC\;\; and \;\; \bigtriangleup BAD$,
AB = AB (Common)
DBA=CBA$\angle DBA =\angle CBA$
Thus, ABCBAD$\bigtriangleup ABC \cong \bigtriangleup BAD$ (by SAS congruence condition).

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

Exercise 2:

Q1. ABCD$ABCD$ is a quadrilateral in which P, Q, R and S are mid-points of the sides AB,BC,CDandDA$AB,BC,CD\; and \;DA$. AC is a diagonal. Show that :
(i) SRAC$SR\parallel AC$ and SR=12AC$SR=\frac{1}{2}AC$
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Solution:

(i) In DAC$\bigtriangleup DAC$,
R is the mid point of DC and S is the mid point of DA.
Thus by mid point theorem, SRAC$SR\parallel AC$ and SR=12AC$SR=\frac{1}{2}AC$

(ii) In BAC$\bigtriangleup BAC$,
P is the mid point of AB and Q is the mid point of BC.
Thus by mid point theorem, PQAC$PQ\parallel AC$ and PQ=12AC$PQ=\frac{1}{2} AC$
also, SR=12AC$SR=\frac{1}{2} AC$
Thus, PQ = SR

(iii) SRAC$SR\parallel AC$– from (i)
and, PQAC$PQ\parallel AC$– from (ii)
SRPQ$\Rightarrow SR \parallel PQ$– 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 DRSandBPQ$\bigtriangleup DRS \;\;and\;\;\bigtriangleup BPQ$,
DS = BQ (Halves of the opposite sides of the rhombus)
SDR=QBP$\angle SDR=\angle QBP$ (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
Thus, DRSBPQ$\bigtriangleup DRS \cong \bigtriangleup BPQ$ (by SAS congruence condition).
RS = PQ (by CPCT)    ………… (i)
In QCRandSAP$\bigtriangleup QCR \;\; and \;\; \bigtriangleup SAP$,
RC = PA (Halves of the opposite sides of the rhombus)
RCQ=PAS$\angle RCQ=\angle PAS$ (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
Thus, QCRSAP$\bigtriangleup QCR \cong \bigtriangleup SAP$ (by SAS congruence condition).
RQ = SP (by CPCT)    ……(ii)
Now,
In CBD$\bigtriangleup CBD$,
R and Q are the mid points of CD and BC respectively.
QRBD$\Rightarrow QR \parallel BD$
also,
P and S are the mid points of AD and AB respectively.
PSBD$\Rightarrow PS \parallel BD$
QRPS$\Rightarrow QR \parallel PS$
Thus, PQRS is a parallelogram.
also, PQR=90$\angle PQR=90^{\circ}$
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
Q=90$\angle Q=90^{\circ}$
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 ABC$\bigtriangleup ABC$
P and Q are the mid-points of AB and BC respectively
Thus, PQAC$PQ\parallel AC$ and PQ=12AC$PQ= \frac{1}{2}AC$ (Mid point theorem) ……….(i)
In ADC$\bigtriangleup ADC$,
SRAC$SR\parallel AC$ and SR=12AC$SR=\frac{1}{2}AC$ (Mid point theorem)   ……(ii)
So, PQSR$PQ\parallel SR$ and PQ=SR$PQ= SR$
As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.
PSQR$PS\parallel QR$ and PS=QR$PS= QR$  (Opposite sides of parallelogram)  ….(iii)
Now,
In BCD$\bigtriangleup BCD$,
Q and R are mid points of side BC and CD respectively.
Thus, QRBDandQR=12BD$QR\parallel BD\;\; and\;\; QR=\frac{1}{2}BD$ (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 ABDC$AB\parallel DC$, 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 ABDC$AB\parallel DC$, 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 BAD$\bigtriangleup BAD$,
E is the mid point of AD and also EGAB$EG\parallel AB$.
Thus, G is the mid point of BD (Converse of mid point theorem)
Now,
In BDC$\bigtriangleup BDC$,
G is the mid point of BD and also GFABDC$GF\parallel AB\parallel DC$.
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, ABCD$AB\parallel CD$

also, AEFC$AE\parallel FC$

Now,
AB=CD$AB=CD$  (Opposite sides of parallelogram ABCD)
12AB=12CD$\Rightarrow \frac{1}{2}AB=\frac{1}{2}CD$
AE=FC$\Rightarrow AE=FC$ (E and F are midpoints of side AB and CD)
AECF is a parallelogram (AE and CF are parallel and equal to each other)
AFEC$AF\parallel EC$ (Opposite sides of a parallelogram)
Now,
In DQC$\bigtriangleup DQC$,
F is mid point of side DC and FPCQ$FP\parallel CQ$ (as AFEC$AF\parallel EC$).
P is the mid-point of DQ (Converse of mid-point theorem)
DP=PQ$\Rightarrow DP=PQ$   ………(i)
Similarly,
In APB$\bigtriangleup APB$,
E is mid point of side AB and EQAP$EQ\parallel AP$ (as AFEC$AF\parallel EC$).
Q is the mid-point of PB (Converse of mid-point theorem)
PQ=QB$\Rightarrow PQ=QB$    …….(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 ACD$\bigtriangleup ACD$,
R and S are the mid points of CD and DA respectively.
Thus, SRAC$SR\parallel AC$.
Similarly we can show that,
PQAC$PQ\parallel AC$
PSBD$PS\parallel BD$
QRBD$QR\parallel BD$
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) MDAC$MD\perp AC$
(iii) CM=MA=12AB$CM=MA=\frac{1}{2}AB$

Solution:
(i) In ACB$\bigtriangleup ACB$,
M is the mid point of AB and MDBC$MD\parallel BC$
Thus, D is the mid point of AC (Converse of mid point theorem)

(ii) ACB=ADM$\angle ACB=\angle ADM$ (Corresponding angles)
also, ACB=90$\angle ACB=90^{\circ}$
Thus, ADM=90$\angle ADM=90^{\circ}$and MDAC$MD\perp AC$

(iii)  In AMDandCMD$\bigtriangleup AMD\;\; and\;\; \bigtriangleup CMD$,
AD = CD (D is the midpoint of side AC)
ADM=CDM$\angle ADM=\angle CDM$ (Each 90°)
DM = DM (common)
Thus, AMDCMD$\bigtriangleup AMD \cong \bigtriangleup CMD$ (by SAS congruence condition).
AM = CM (by CPCT)
also, AM=12AB$AM=\frac{1}{2}AB$ (as M is mid point of AB)
Hence, CM=AM=12AB$CM=AM=\frac{1}{2}AB$.