# NCERT solution for class 9 maths areas of parallelograms and triangles ( Chapter 9)

#### Solution for Exercise 9.1

1.Which of the following figures lie on the same base and between the same parallels.
In such a case, write the common base and the two parallels.
i)ii)iii)
iv)v)vi)

i)
Yes. It can be observed that trapezium ABCD and triangle PCD have a common base CD and these are lying between the same parallel lines AB and CD.

ii)
No. It can be observed that parallelogram PQRS and trapezium MNRS have a common base RS.
However, their vertices, (i.e., opposite to the common base) P, Q of parallelogram and M, N of trapezium, are not lying on the same line.

iii)
Yes. It can be observed that parallelogram PQRS and triangle TQR have a common base QR and they are lying between the same parallel lines PS and QR.

iv)
No. It can be observed that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and ac. However, these do not have any common base.

v)
Yes. It can be observed that parallelogram ABCD and parallelogram APQD have a common base AD and these are lying between the same parallel lines AD and BQ.

vi)
No. It can be observed that parallelogram PBCS and PQRS are lying on the same base PS.
However, these do not lie between the same parallel lines.

#### Solution for Exercise 9.2

1. In Figure, ABCD is a parallelogram, $$AE\perp{DC}$$ and $$CF\perp{AD}$$.
If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

In parallelogram ABCD, CD = AB = 16 cm [Opposite sides of a parallelogram are equal]
We know that,
Area of a parallelogram = Base x Corresponding altitude
i.e., Area of parallelogram ABCD = CD x AE ...(i)
Also, Area of parallelogram = Base x Corresponding altitude
Area of parallelogram ABCD = AD x CF ...(ii)
Thus, from (i) and (ii), we get,
CD x AE = AD x CF
i.e., 16cm x 8cm = AD x 10cm
i.e., AD = 16cm x 8cm/10cm
Therefore, the length of AD is 12.8 cm.

2. If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar (ABCD)

Given: E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD

Construction: Let us join HF.

To prove: ar(EFGH) = 1/2 ar(ABCD)

Proof: In parallelogram ABCD,
AD = BC and AD || BC ( Opposite sides of a parallelogram are equal and parallel)
Multiplying both sides by 1/2, we get,
i.e., 1/2 x AD = 1/2 x BC
i.e., AH = BF and AH || BF
Therefore, ABFH is a parallelogram.

Since, $$\triangle{HEF}$$ and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,
Thus, Area ($$\triangle{HEF}$$) = 1/2 Area (ABFH) ...(i)
Similarly, it can be proved that,
Area ($$\triangle{HGF}$$) = 1/2 Area (HDCF) ...(ii)
On adding equations (i) and (ii), we get,
i.e., Area ($$\triangle{HEF}$$) + Area ($$\triangle{HGF}$$) = 1/2 Area (ABFH) + 1/2 Area (HDCF)
i.e.Area [($$\triangle{HEF}$$) + ($$\triangle{HGF}$$)] = 1/2 [Area (ABFH) + Area (HDCF)]
Therefore, Area (EFGH) = 1/2 Area (ABCD)
Hence, proved.

3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.
Show that ar(APB) = ar(BQC).

It can be observed that $$\triangle{BQC}$$ and paralleogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC.

Therefore, Area($$\triangle{BQC}$$) = 1/2 Area(ABCD) ...(i)
Similarly, $$\triangle{APB}$$ and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC.
Therefore, Area($$\triangle{APB}$$) = 1/2 Area(ABCD) ...(ii)
From eq. (i) and (ii), we get,
Area($$\triangle{BQC}$$) = Area($$\triangle{APB}$$)
Hence, proved.

4. In Figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar ($$\triangle{APB}$$ + $$\triangle{PCD}$$) = 1/2 ar (ABCD)
(ii) ar ($$\triangle{APD}$$ + $$\triangle{PBC}$$) = ar(APB) + ar(PCD)

[Hint : Through P, draw a line parallel to AB.]

Construction: Let us draw a line segment EF, passing through point P and parallel to line segment AB.

In parallelogram ABCD, AB || EF ...(i)(By construction)
Also, ABCD is a parallelogram
So, AD || BC ...(Opposite sides of parallelogram)
Also, AE || BF ...(ii)(Since, 1/2 AD = AE and 1/2 BC = BF)
From equations (i) and (ii), we get,
AB || EF and AE || BF
Therefore, quadrilateral ABFE is a parallelogram.

It can be observed that $$\triangle{APB}$$ and parallelogram ABFE are lying on the same base AB and between the same parallel lines AB and EF.
Therefore, we can say that,
Area $$\triangle{APB}$$ = 1/2 Area (ABFE) ...(iii)
Similarly, for $$\triangle{PCD}$$ and parallelogram EFCD,
i.e., Area $$\triangle{PCD}$$ = 1/2 Area (EFCD) ...(iv)
Now, adding eq. (iii) and (iv), we get,
Area $$\triangle{APB}$$ + Area $$\triangle{PCD}$$ = 1/2 Area (ABFE) + 1/2 Area (EFCD)
i.e., Area $$\triangle{APB}$$ + Area $$\triangle{PCD}$$ = 1/2 [Area (ABFE) + Area (EFCD)]
i.e., Area $$\triangle{APB}$$ + Area $$\triangle{PCD}$$ = 1/2 Area (ABCD) ...(v)
Hence, part i) is proved.

Construction: Let us draw a line segment MN, passing through point P and parallel to line segment AD.

In parallelogram ABCD, MN || AD ...(vi)(By construction)
Also, ABCD is a parallelogram
Thus, AB || DC ...(Opposite sides of parallelogram)
Also, AM || DN ...(vii)(Since, 1/2 AB = AM and 1/2 DC = DN)
From equations (vi) and (vii), we get,
MN || AD and AM || DN
Therefore, quadrilateral AMND is a parallelogram.
It can be observed that $$\triangle{APD}$$ and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN.
Therefore, we can say that,
Area $$\triangle{APD}$$ = 1/2 Area (AMND) ...(viii)

Similarly, for $$\triangle{PCB}$$ and parallelogram MNCB,
i.e., Area $$\triangle{PCB}$$ = 1/2 Area (MNCB) ...(ix)
Now, adding eq. (viii) and (ix), we get,
Area $$\triangle{APD}$$ + Area $$\triangle{PCB}$$ = 1/2 Area (AMND) + 1/2 Area (MNCB)
i.e., Area $$\triangle{APD}$$ + Area $$\triangle{PCB}$$ = 1/2 [Area (AMND) + Area (MNCB)]
i.e., Area $$\triangle{APD}$$ + Area $$\triangle{PCB}$$ = 1/2 Area (ABCD) ...(x)
Thus, comparing eq. (v) and (x), we get,
Area $$\triangle{APB}$$ + Area $$\triangle{PCD}$$ = Area $$\triangle{APD}$$ + Area $$\triangle{PCB}$$
Hence, proved.

5. In Figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
i)ar(PQRS) = ar(ABRS)
ii) ar(AXS) = 1/2 ar(PQRS)

i) It can be observed that paralleogram PQRS and ABRS lie on the same base SR and also, these lie in between the same parallel lines SR and PB.
Therefore, Area (PQRS) = Area (ABRS) ...(i)

ii) Consider $$\triangle{AXS}$$ and paralleogram ABRS.
As these lie on the same base and are between the same parallel lines AS and BR
Therefore, Area ($$\triangle{AXS}$$) = 1/2 Area (ABRS) ...(ii)
From equations (i) and (ii), we get,
ar(AXS) = 1/2 ar(PQRS)
Hence, proved.

6. A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q.
In how many parts the fields is divided? What are the shapes of these parts?
The farmer wants to sow wheat and pulses in equal portions of the field separately.
How should she do it?

From the figure, we can say that it is that point A which divides the field into three parts.
These parts are triangular in shape i.e., $$\triangle{PSA}$$, $$\triangle{PAQ}$$ and $$\triangle{QRA}$$
Thus, Area $$\triangle{PSA}$$ + Area $$\triangle{PAQ}$$ + Area $$\triangle{QRA}$$ = Area of Parallelogram PQRS ...(i)
Also, Area $$\triangle{PAQ}$$ = (1/2) Area of Parallelogram PQRS ...(ii)
From equation (i) and (ii), we get,
Area $$\triangle{PSA}$$ + Area $$\triangle{QRA}$$ = (1/2) Area of Parallelogram PQRS ...(iii)
Clearly, it can be observed that the farmer must sow wheat in triangular part PAQ and pulses in other two triangular parts PSA and QRA or wheat in triangular parts PSA and QRA and pulses in triangular parts PAQ.

#### Solution for Exercise 9.3

1. In Figure, E is any point on median AD of a $$\triangle{ABC}$$. Show that ar (ABE) = ar (ACE)

AD is the median of $$\triangle{ABC}$$. Therefore, it will divide $$\triangle{ABC}$$ in to two triangles of equal areas.
Thus, Area ($$\triangle{ABD}$$) = Area ($$\triangle{ACD}$$) ...(i)
Similarly, ED is the median of $$\triangle{EBC}$$
Thus, Area ($$\triangle{EBD}$$) = Area ($$\triangle{ECD}$$) ...(ii)
On subtracting equation (ii) from equations (i), we get,
i.e., Area ($$\triangle{ABD}$$) - Area ($$\triangle{EBD}$$) = Area ($$\triangle{ACD}$$) - ($$\triangle{ECD}$$)
Therefore, Area ($$\triangle{ABE}$$) = Area ($$\triangle{AEC}$$)
Hence, proved.

2. In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4 ar (ABC).

In $$\triangle{BED}$$,
ar ($$\triangle{BED}$$) = 1/2 × BD × DE
Also, as E is the mid-point of AD
We get, AE = DE
As AD is the median on side BC of triangle ABC,
Therefore, BD = DC

and BD = (1/2) BC ...(ii)
From (i) and (ii), we get,
ar(BED) = (1/2) × (1/2)BC × (1/2)AD
i.e., ar(BED) = (1/2) × (1/2) ar(ABC)
i.e., ar(BED) = 1/4 ar(ABC)
Hence, proved.

3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.

We know that, diagonals of parallelogram bisect each other.
Therefore, O is the mid-point of AC and BD.

Now, in $$\triangle{ABC}$$ ,BO is the median.
Therefore, it will divide it into two triangles of equal areas.
Thus, Area ($$\triangle{AOB}$$) = Area ($$\triangle{BOC}$$) ...(i)
Similarly, in $$\triangle{BCD}$$ ,BO is the median.
Therefore, Area ($$\triangle{BOC}$$) = Area ($$\triangle{COD}$$) ...(ii)
Similarly, Area ($$\triangle{COD}$$) = Area ($$\triangle{AOD}$$) ...(iii)

Thus, from equations (i), (ii), and (iii), we get,
Area ($$\triangle{AOB}$$) = Area ($$\triangle{BOC}$$) = Area ($$\triangle{COD}$$) = Area ($$\triangle{AOD}$$)
Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of equal area.
Hence, proved.

4. In Figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD).

In $$\triangle{ACD}$$,
Line-segment CD is bisected by AB at O.
So, AO is the median of $$\triangle{ACD}$$.
Therefore, Area ($$\triangle{ACO}$$) = Area ($$\triangle{ADO}$$) ...(i)
Similarly, in $$\triangle{BCD}$$, BO is the median.
Therefore, Area ($$\triangle{BCO}$$) = Area ($$\triangle{BDO}$$) ...(ii)
Now, adding equations (i) and (ii), we get,
Area ($$\triangle{ACO}$$) + Area ($$\triangle{BCO}$$) = Area ($$\triangle{ADO}$$) + Area ($$\triangle{BDO}$$)
Therefore, Area ($$\triangle{ABC}$$) = Area ($$\triangle{ABD}$$)
Hence, proved.

5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a $$\triangle{ABC}$$. Show that
i) BDEF is a parallelogram.
ii) ar(DEF) = 1/4 ar(ABC)
iii) ar(BDEF) = 1/2 ar(ABC)

i) In $$\triangle{ABC}$$,
EF || BC and EF = 1/2 BC ...(By mid point theorem)
Also, we have,
BD = 1/2 BC ...(Since, D is the mid point)
So, we get, BD = EF
Also, BF and DE will be parallel and equal to each other.
Thus, the pair opposite sides are equal in length and parallel to each other.
Therefore, BDEF is a parallelogram.

ii) Proceeding from the results of i)
We get, BDEF, DCEF, AFDE are parallelograms
Diagonal of a parallelogram divides it into two triangles of equal area.
Therefore, ar($$\triangle{BFD}$$) = ar($$\triangle{DEF}$$) ...(i)(From parallelogram BDEF)
Also, ar($$\triangle{AFE}$$) = ar($$\triangle{DEF}$$) ...(ii)(From parallelogram DCEF)
Similarly, ar($$\triangle{CDE}$$) = ar($$\triangle{DEF}$$) ...(iii)(From parallelogram AFDE)
Thus, from eq. (i), (ii) and (iii), we get,
ar($$\triangle{BFD}$$) = ar($$\triangle{AFE}$$) = ar($$\triangle{CDE}$$) = ar($$\triangle{DEF}$$)
ar($$\triangle{BFD}$$) + ar($$\triangle{AFE}$$) + ar($$\triangle{CDE}$$) + ar($$\triangle{DEF}$$) = ar($$\triangle{ABC}$$)
4 ar($$\triangle{DEF}$$) = ar($$\triangle{ABC}$$)
Therefore, ar($$\triangle{DEF}$$) = 1/4 ar($$\triangle{ABC}$$)

ar (parallelogram BDEF) = ar ($$\triangle{DEF}$$) + ar($$\triangle{BDE}$$)
= ar (parallelogram BDEF) = ar ($$\triangle{DEF}$$) + ar($$\triangle{DEF}$$)
= ar (parallelogram BDEF) = 2 x 1/4 ar ($$\triangle{ABC}$$)
Therefore, ar (parallelogram BDEF) = 1/2 ar ($$\triangle{ABC}$$)
Hence, proved.

6. In Figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
i) ar (DOC) = ar (AOB)
ii) ar (DCB) = ar (ACB)
iii) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]

Construction: Let us draw $$DN\perp{AC}$$ and $$BN\perp{AC}$$.

i) In ($$\triangle{DON}$$) and ($$\triangle{BOM}$$),
$$\angle{DNO}$$ = $$\angle{BMO}$$ ...(By construction)
$$\angle{DON}$$ = $$\angle{BOM}$$ ...(Vertically opposite angles)
Also, OD = OB ...(Given)
Therefore, $$\triangle{DON}$$ $$\displaystyle \cong$$ $$\triangle{BOM}$$ ...(By AAS congruency test)
Thus, DN = BM ...(i)(CPCT)
But, we know that, congruent triangles have equal areas.
Therefore, Area ($$\triangle{DON}$$) = Area ($$\triangle{BOM}$$) ...(ii)

Now, in ($$\triangle{DNC}$$) and ($$\triangle{BMA}$$),
$$\angle{DNC}$$ = $$\angle{BMA}$$ ...(By construction)
DN = BM ...(Using Equation (i))
Also, CD = AB ...(Given)
Therefore, $$\triangle{DNC}$$ $$\displaystyle \cong$$ $$\triangle{BMA}$$ ...(By RHS congruency test)
Therefore, Area ($$\triangle{DNC}$$) = Area ($$\triangle{BMA}$$) ...(iii)
On adding eq. (ii)and (iii), we get,
Area $$\triangle{DON}$$ + Area $$\triangle{DNC}$$ = Area $$\triangle{BOM}$$ + Area $$\triangle{BMA}$$
Therefore, Area $$\triangle{DOC}$$ = Area $$\triangle{AOB}$$

ii)We have, Area $$\triangle{DOC}$$ = Area $$\triangle{AOB}$$
Adding Area $$\triangle{OCB}$$ on both sides, we get,
i.e., Area $$\triangle{DOC}$$ + Area $$\triangle{OCB}$$ = Area $$\triangle{AOB}$$ + Area $$\triangle{OCB}$$
i.e., Area $$\triangle{DCB}$$ = Area $$\triangle{ACB}$$

iii)Now, we have, Area $$\triangle{DCB}$$ = Area $$\triangle{ACB}$$
We also know that, if two triangles have the same base and equal areas, then these will lie between the same parallels.
i.e., DA || CB
In quadrilateral ABCD, one pair of opposite sides is equal (AB = CD ) and the other part of opposite sides is parallel (DA || CB)
Therefore, we can say that, ABCD is a parallelogram.
Hence, proved.

7. D and E are points on sides AB and AC respectively of $$\triangle{ABC}$$ such that ar (DBC) = ar (EBC). Prove that DE || BC.

Since $$\triangle{BCE}$$ and $$\triangle{BCD}$$ are lying on a common base BC and also have equal areas.
Therefore, $$\triangle{BCE}$$ and $$\triangle{BCD}$$ will lie between the same parallel lines.
i.e., DE || BC
Hence, it is proved that DE || BC

8. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively,
show that ar (ABE) = ar (ACF)

XY || BC = EY || BC ...(Given)
Also, BE || AC = BE || CY
Therefore, we can say that, EBYC is a parallelogram.
Similarly, XY || BC = XF || BC
Also, FC || AB = FC || XB
Therefore, BCFX is a parallelogram.

Parallelograms EBCY and BCFX are on the same base BC and between the same parallels BC and EF.
Thus, Area (EBCY) = (1/2) Area (BCFX) ...(i)
Now, considering parallelogram EBYC and $$\triangle{AEB}$$,
We can say that, these lie on the same base BE and are between the same parallels BE and AC.
Thus, Area ($$\triangle{AEB}$$) = (1/2) Area (EBYC) ...(ii)
Also, parallelogram BCFX and $$\triangle{ACF}$$ are on the same base CF and between the same parallels CF and AB.
Thus, Area ($$\triangle{ACF}$$) = (1/2) Area (BCFX) ...(iii)
So, from eq.(i), (ii) and (iii), we get,
Area ($$\triangle{AEB}$$) = ($$\triangle{ACF}$$)
Hence, proved.

9. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Figure).
Show that ar (ABCD) = ar (PBQR).
[Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]

Construction: Let us join AC and PQ.

$$\triangle{ACQ}$$ and $$\triangle{AQP}$$ are the same base AQ and between the same parallels AQ and CP.
Therefore, Area $$\triangle{ACQ}$$ = Area $$\triangle{AQP}$$
Area ($$\triangle{ACQ}$$) - Area ($$\triangle{ABQ}$$) = Area ($$\triangle{APQ}$$) - Area ($$\triangle{ABQ}$$)
Thus, ($$\triangle{ABC}$$) = Area ($$\triangle{QBP}$$) ...(i)

Since, AC and PQ are diagonals of parallelograms ABCD and PBQR, respectively
Therefore, Area ($$\triangle{ABC}$$) = (1/2) Area (ABCD) ...(ii)
Also, Area ($$\triangle{QBP}$$) = (1/2) Area (PBQR) ...(iii)
From equations (i),(ii) and (iii), we get,
(1/2) Area (ABCD) = (1/2) Area(PBQR)
Therefore, Area (ABCD) = Area(PBQR)
Hence, proved.

10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

It can be observed that $$\triangle{DAC}$$ and $$\triangle{DBC}$$ lie on the same base DC and between the same parallels AB and CD.
Therefore, Area ($$\triangle{DAC}$$) = Area ($$\triangle{DBC}$$)
i.e., Area ($$\triangle{DAC}$$) - Area ($$\triangle{DOC}$$) = Area ($$\triangle{DBC}$$) - Area ($$\triangle{DOC}$$)
Area ($$\triangle{AOD}$$) = Area ($$\triangle{BOC}$$)
Hence, proved.

11. In Figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
i) ar (ACB) = ar (ACF)
ii) ar (AEDF) = ar (ABCDE).

i) $$\triangle{ACB}$$ and $$\triangle{ACF}$$ lie on the same base AC and are between The same parallels AC and BF.
Therefore, Area ($$\triangle{ACB}$$) = Area ($$\triangle{ACF}$$)

ii) Since, it can be observed that
Area ($$\triangle{ACB}$$) = Area ($$\triangle{ACF}$$)
Thus, Area ($$\triangle{ACB}$$) + Area (ACDE) = Area ($$\triangle{ACF}$$) + Area (ACDE)
Therefore, Area (ABCDE) = Area (AEDF)
Hence, proved.

12. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre.
Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot.
Explain how this proposal will be implemented.

Let quadrilateral ABCD be the original shape of the field.

The proposal may be implemented as follows:
Join diagonal BD and draw a line parallel to BD through point A.
Let it meet the extended side CD of ABCD at point E.
Join BE and AD. Let them intersect each other at O.
Then, portion $$\triangle{AOB}$$ can be cut from the original field so that the new shape of the field will be $$\triangle{BCE}$$ as shown below.

We have to prove that the area of $$\triangle{AOB}$$ (portion that was cut so as to construct Health Centre) is equal to the area of $$\triangle{DEO}$$(portion added to the field so as to make the area of the new field so formed equal to the area of the original field).
It can be observed that $$\triangle{DEB}$$ and $$\triangle{DAB}$$ lie on the same base BD and are between the same parallels BD and AE.
Area ($$\triangle{DEB}$$) = Area ($$\triangle{DAB}$$)
Area ($$\triangle{DEB}$$) - Area ($$\triangle{DOB}$$) = ($$\triangle{DAB}$$) - Area ($$\triangle{DOB}$$)
Therefore, Area ($$\triangle{DEO}$$) = Area ($$\triangle{AOB}$$)

13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y.
Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]

It can be observed that, $$\triangle{ADX}$$ and $$\triangle{ACX}$$ lie on the same base AX and are between the same parallels AB and DC.
Therefore, Area ($$\triangle{ADX}$$) = Area ($$\triangle{ACX}$$) ...(i)
Also, $$\triangle{ADY}$$ and $$\triangle{ACX}$$ lie on the same base AC and are between the same parallels AC and XY.
Therefore, Area ($$\triangle{ACY}$$) = Arca ($$\triangle{ACX}$$) ...(ii)
From Equations (i) and (ii), we get,
Area ($$\triangle{ADX}$$) = Area ($$\triangle{ACY}$$)
Hence, proved.

14. In Figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

It can be observed that, $$\triangle{ABQ}$$ and $$\triangle{PBQ}$$ lie on the same base BQ and are between the same parallels AP and BQ.
Therefore, Area ($$\triangle{ABQ}$$) = Area ($$\triangle{PBQ}$$) ...(i)
Also, $$\triangle{BCQ}$$ and $$\triangle{BRQ}$$ lie on the same BQ and are between the same parallels BQ and CR.
Therefore, Area ($$\triangle{BCQ}$$) = Arca ($$\triangle{BRQ}$$) ...(ii)
From Equations (i) and (ii), we get,
i.e.,Area ($$\triangle{ABQ}$$) + Area($$\triangle{BCQ}$$) = Area ($$\triangle{PBQ}$$) + Area ($$\triangle{BRQ}$$)
Area ($$\triangle{AQC}$$) = Area ($$\triangle{PBR}$$)
Hence, proved.

15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC).
Prove that ABCD is a trapezium.

Given:
Area ($$\triangle{AOD}$$) = Area ($$\triangle{BOC}$$)
i.e., Area ($$\triangle{AOD}$$) + Area ($$\triangle{AOB}$$) = ($$\triangle{BOC}$$) + Area ($$\triangle{AOB}$$)
Therefore, Area ($$\triangle{ADB}$$) = Area ($$\triangle{ACB}$$)
We know that triangles on the same base having area equal to each other lie between the same parallels.
Therefore, these triangles, $$\triangle{ADB}$$ and $$\triangle{ACB}$$, are lying between the same parallels.i.e., AB || CD
Therefore, ABCD is a trapezium.
Hence, proved.

In Figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
Given: Area ($$\triangle{DRC}$$) = Area ($$\triangle{DPC}$$).
As, $$\triangle{DRC}$$ and $$\triangle{DPC}$$ lie on the same base DC and have equal area, therefore, they must lie between the same parallel lines.
Also, Area ($$\triangle{BDP}$$) = Area ($$\triangle{ARC}$$).
i.e., Area ($$\triangle{BDP}$$) - Area ($$\triangle{DPC}$$) = Area ($$\triangle{ARC}$$) - Area ($$\triangle{DRC}$$)
Therefore, Area ($$\triangle{BDC}$$) = Area($$\triangle{ADC}$$)
Since, $$\triangle{BDC}$$ and $$\triangle{ADC}$$ lie on the same base CD and have equal areas, they must lie between the same parallel lines.