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SEBA Class 9 Mathematics Chapter 9 Areas of Parallelograms and Triangle
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Areas of Parallelograms and Triangle
Chapter – 9
| 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.
Ans: (i) ∆PDC and quadrilateral ABCD lie on the same base DC and between the same parallels DC and AB.
(iii) ∆TRQ and parallelogram SRQP lie on the same base RQ and between the same parallels RQ and SP.
(v) Quadrilaterals ABCD and ABQD lie on the same base AD and between the same parallels AD and BQ.
| IMPORTANT POINTS |
1. Theorem: Parallelograms on the same base and between the same parallels are equal in area.
Given: Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF and DC.
To Prove: ar (ABCD) = ar (EFCD).
Proof: In ∆ ADE and ∆ BCF,
∠DAE = ∠CBF …(1) | Corresponding angles from AD || BC and transversal AF
∠AED = ∠BFC …(2) | Corresponding angles from ED || FC and transversal AF
∴ ∠ADE = ∠ BCF …(3) | Angle sum property of a triangle
Also, AD = BC …(4) | Opposite sides of the parallelogram ABCD
From (1), (3) and (4) we get
∆ ADE ≅ ∆ CBF | By ASA Rule
∴ ar (∆ ADE) = ar (∆BCF) … (5) | ∴ Congruent figures have equal areas
Now, ar (ABCD) = ar (ADE) + ar (EDCB) | From (5)
= ar (BCF)+ar (EDCB) = ar (EDCB)
So, parallelograms ABCD and EFCD are equal in areas.
2. The area of a parallelogram is the product of its any side and the corresponding altitude.
Arca of the parallelogram ABCD = AB × DE
3. If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.
| Exercise 9.2 |
1. In figure, ABCD is a parallelogram, AE⊥DC, and CF⊥AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm find AD.
Ans: ar (parallelogram ABCD) = AB X AE
= 16 x 8 cm²
= 128 cm² … (1)
ar (parallelogram ABCD) = AD X CF
= AD x 10 cm² … (2)
From (1) and (2), we get
AD x 10 = 128
2. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar 
Ans: Given: E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD.
Construction: Join OF, OG, OH and OE. Also, join AC and BD.
Proof: In ∆BCD,
∵ Fand G are the mid-points of BC and DC respectively.
∴ | FG || BD …(1) In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side
In ABAD,
∵ E and H are the mid-points of AB and AD respectively.
∴ EH || BD..(2) | In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side
From (1) and (2),
EH || BD …(3)
Similarly, we can prove that
EF || HG …(4)
From (3) and (4),
Quadrilateral EFGH is a parallelogram | A quadrilateral is a parallelogram if its opposite sides are equal
∴ F is the mid-point of CB and O is the point of CA
∴ FO || BA In a triangle, the line segment joining the mid- points of any two sides is parallel to the third side and is half of it
⇒ FO || CG … (5)
∵ BA || CD (opposite sides of a parallelogram are parallel)
∴ BA || CG
From (5) and (6) we get
Quadrilateral OFCG is a parallelogram
∵ A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length
∴ OP = PC | ∵ Diagonals of a || gm bisect each other
∵ ∆OPF and ∆ CPF have equal bases (∵ OP = PC) and have a common vertex F
∴ Their altitudes are also the same
∴ ar (A OPF) = ar (∆ CDF)
Similarly, ar (∆ OQF) = ar (∆ BQF)
Adding, we get
ar (∆ OPF) + ar (∆ OQF) = ar (∆CPF) + ar (∆ BQF)
⇒ ar (|| gm OQFP) = ar (∆ CPF) + ar (∆BQF) …(7)
Similarly, ar (|| gm OPGS) = ar (∆ GPC) + ar (∆ DSG) …(8)
ar (|| gm OSHR) = ar (∆DSH) + ar (∆ HAR) …(9)
ar (|| gm OREQ) = ar (∆ ARE) + ar (∆ EQB)…(10)
Adding the corresponding sides of (7), (8), (9) and (10), we get
ar (|| gm EFGH) = {ar (∆ CPF) + ar (∆ GPC) + {ar (∆ DSG) + ar (∆DSH)} + {ar (∆HAR) + ar (∆ARE)} + {ar (∆ BQF) + (∆EQB)}
= ar (∆ FCG) + ar (∆GDH) + ar (∆ HAE ) +ar (∆EBF)
3. P and Q are any two points lying on the sides DC and AD respec-tively of a parallelogram ABCD. Show that ar(APW) = ar(BQC)
Ans: Given: P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.
To Prove: ar(APW) = ar(BQC)
Proof: ∵ ∆ APB and || gm ABCD are on the same base AB and between the same parallels AB and DC.
∵ BQC and || gm ABCD are on the same base BC and between the same parallels BC and AD.
From (1) and (2),
ar(∆APB) = ar(∆BQC)
4. In figure, P is a point in the interior of a parallelogram ABCD. Show that:
(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
[Hint. Through P, draw a line parallel to AB]
Ans: Given: P is a point in the interior of a parallelogram ABCD.
(ii) ar(∆APD) + ar(∆PBC) = ar(∆APB) + ar(∆PCD)
Construction: Through P, draw a line EF parallel to AB.
Proof: (i) EF || AB …(1) | by construction
∵ AD ||BC | ∵ Opposite sides of a parallelogram are parallel
∴ AE|| BF …(2)
From (1) and (2) we get
Quadrilateral ABFE is a parallelogram | A quadrilateral is a parallelogram if its opposite sides are parallel Similarly, quadrilateral CDEF is a parallelogram
∵ ∆APB and || gm ABFE are on the same base AB and between the same parallels AB and EF
…(3)
∴ ∆ PCD and || gm CDEF are on the same base DC and between the same parallels DC and EF.
…(4)
(3) + (4) ⇒
(ii) ar(∆APD) + ar(∆PBC)
= ar (ll gm ABCD) – [ar(∆APB)+ar(PCD)]
= 2[ar (∆ APB)+ar( ∆ PCB)] – [ar(∆APB) + ar(∆PCD)]
= ar(∆APB) + ar(∆PCD)
5. In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that:
(i) ar (PQRS) = ar(APRS)
Ans: Given: PQRS and ABRS parallelograms and X is any point on side BR.
To Prove: (i) ar (PQRS) = ar(ABRS)
Proof: (i) In∆PSA and ∆ QRB
∠ SPA =∠RQB …(1)
∠ PAS = ∠QBR …(2) | Corresponding angles from PS || QR and transversal PB Corresponding angles from SS || BR and transversal PB
∴ ∠ PSA =∠QRB …(3) | Angle sum property of a triangle
Also, PS = QR …(4) | Opposite sides of || gm PQRS
From (1), (3) and (4) we get
∆ PSA ≅ ∆ QRB …(5) | By ASA Rule
∴ ar(∆PSA) = ar(∆QRB) …(6) ∵ Congruent figures have equal areas
Now, ar(PQRS) = ar(∆PSA) + ar(AQRS)
= ar(∆QRB) + ar(AQRS) | Using (6)
= ar (ABRS)
(ii) ∵ ∆ AXS and || gm ABRS are on the same base AS and between the same parallels AS and BR.
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 is the field divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field sepa-rately. How should she do it?
Ans: The field is divided into three parts each of the shape of a triangle.
There are:
(i) ∆ APS
(ii) ∆ APQ
(iii) ∆ AQR
If the farmer want to sow wheat and pulses in equal portions of the field separately, then A must be taken
SR || PQ | ∵ PQRS is a parallelogram
∴ AR || PB …(1)
SR = PQ | ∵ Opposite sides of a parallelogram are equal
∴ AR = PB …(2)
From (1) and (2) we get
APBP is a parallelogram. | A quadrilateral is a || gm if its one pair of opposite sides is parallel and of equal length.
∵ ∆ ASP, || gm ARBP and ∆ BQR are in between the same parallels PQ and SR.
∴ Their altitudes are equal. Let it be x.
| by assumption
| by assumption
…(5)
From (3), (4) and (5) we get
ar (∆ASP) = ar (ll gm ARBP) = ar (∆BQR)
| Exercise 9.3 |
1. In figure, E is any point on the median AD of a ∆ABC. Show that ar(ABE) = ar(ACE).
Ans: Given: E is any point on the median AD of ∆ ABC.
To Prove: ar(∆ABC) = ar(∆ACE).
Proof: In ∆ABC,
∵ AD is a median.
∴ ar (∆ABD) = ar(∆ACD) …(1) | ∵ A median of a triangle divides it into two triangles of equal areas
In ∆EBC,
∵ ED is a median.
∴ ar(∆EBD) = ar(∆ECD) …(2)
∵ A median of a triangle divides it into two triangles of equal areas
Subtracting (2) from (1), we get
ar (∆ABD) – ar(∆EBD) = ar(∆ACD) – ar(∆ECD)
⇒ ar(∆ABE) = ar(∆ACE)
2. In a triangle ABC, E is the midpoint of median AD. Show that
Ans: Given: In a triangle ABC, E is the midpoint of median AD.
Proof: In ∆ABC
∵ AD is a median.
In A ABD,
∵ BE is a median.
| ∵ A median of a triangle divides it into two triangles of equal areas
3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Ans: Given: ABCD is a parallelo-gram whose diagonals AC and BD in-tersecting at O divide it into four tri-angle ΔΟΑВ, ∆ОВС, ∆ОСA and ∆OAD.
To Prove: ar (∆OAB) = ar (∆OBC) = ar (ΔOCD) = ar (∆ODA)
Construction: Draw BE⊥AC
Proof: ∵ ABCD is a parallelogram
∴ OA = OC and OB = OD ∵ Diagonals of a parallelogram bisect each other
But OA = OC
∴ ar (∆OAB) = ar (∆OBC) …(1)
Similarly,
ar (∆OBC) = ar (∆OCD) …(2)
and ar (∆OCD) = ar (∆ODA) …(3)
From (1), (2) and (3),
ar (ΔΟΑΒ) = ar (∆OBC) = ar (∆OCB) = ar (∆ODA)
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).
Ans: Given: ABC and DBC are two triangles on the same base AB. Line segment CD is bisected by AB at O.
To Prove: ar (∆ABC) = ar (∆ABD)
Proof: ∵ Line segment CD is bisected by AB at O.
∴ OC = OD
∴ BO is a median of ∆BCD
and AO is a median of ∆ACD
∵ BO is a median of ∆BCD
∴ ar (AOBC) = ar (∆OBD) …(1) | A median of a triangle divides it into two triangles of equal areas
∵ AO is a median of ∆ACD
∴ ar (∆OAC) = ar (∆OAD) …(2) | A median of a triangle divides it into two triangles of equal areas
Adding (1) and (2), we get
ar (∆OBC) + ar (∆OAC) = ar (∆OBD) + ar ( ∆OAD)
⇒ ar (∆ABC) = ar (∆ABD)
5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ABC. Show that:
(i) BDEF is a parallelogram
Ans: Given: D, E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ABC.
Proof: (i) In ∆ABC,
∵ F is the midpoint of side AB and E is the midpoint of side AC.
∴ EF || BC
∵ In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side.
⇒ EF || BD …(1)
Similarly, ED || BF …(2)
In view of (1) and (2),
| ∵ A quadrilateral is a parallelogram if opposite sides are parallel
(ii) As in (i), we can prove that
∵ FD is a diagonal of || gm BDEF.
∴ ar (∆FBD) = ar (∆DEF) …(3)
Similarly, ar (∆DEF) = ar (∆FAE) …(4)
and, ar (∆DEF) = ar (∆DCE) …(5)
From (3), (4) and (5), we have
ar (∆FBD) = ar (∆DEF) = ar (∆FAE) = ar (∆DCE)
∵ ∆ABC is divided into four non-overlapping triangles.
∆FBD, ∆DEF, ∆FAE and ∆DCE
∴ ar (∆ABC) = ar (∆FBD) + (∆DEF) + ar (FAE) + ar (∆DCE)
= 4ar (∆DEF) | From (6)
…(7)
= ar (ΔFBD) + ar ( ∆DEF)
= ar (∆DEF) + ar (∆DEF) | From (3)
= 2 ar (∆DEF)
| From (7)
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 C]
Ans: Given: Diagonals AC and BD of a quadrilateral ABCD intersect at O such that OB = OD.
To Prove: If AB = CD, then
(i) ar (ΔDOC) = ar (ΔΑΟΒ)
(ii) ar (ΔDCB) = ar (ΔACB)
(iii) DA || CB or ΔBCD is a parallelogram
Construction: Draw DE⊥CB and BF⊥ AC
Proof: (iii) In ΔADB,
∵ AO is a median
∴ ar (ΔAOD) = ar (ΔΑΟΒ) …(1) | ∵ A median of a triangle divides it into two triangles of equal areas In ΔCBD,
∴ CO is a median,
∴ ar (ACOD) = ar (ACOB) …(2) | ∵ A median of a triangle divides it into two triangles of equal areas Adding (1) and (2), we get
ar (∆AOD)+ar (∆COD) = ar (∆AOB) + ar (∆COB)
⇒ ar (∆ACD) = ar (∆AСВ)
⇒ DE = BF …(3)
In right ∆S DEF and BFA,
Hyp. DC = Hyp. ВА | Given
DE = BF | From(3)
∴ ∆ DEC ≅ ABFA | R.H.S. Rule
∴ ∠DCE = ∠BAF | c.p.c.t.
But these angles form a pair of equal alternate interior angles.
∴ DC || AB …(4)
∵ DC = AB and DC || AB
| ∵ A quadrilateral is a parallelogram if a pair of opposite sides is parallel and equal
∴ DA || CB | ∵ Opposite sides of a || gm are parallel
(i) ∵ ABCD is a parallelogram
∴ OC = OA …(5)
| ∵ Diagonals of a parallelogram bisect each other
∴ DE = BF | From (3)
and OC = OA | From (5)
∴ ar (∆DOC) = ar (∆AΟΒ)
(ii) From (i), ar (∆DOC) = ar (∆AOB)
⇒ ar (∆DOC) + ar (∆OCB) = ar (∆AOB) + ar (∆OCB) | Adding same areas on both sides
⇒ ar (∆DCB) = ar (∆ACB)
7. D and E are points on sides AB and AC respectively of ∆AB C Such that ar(ADBC) = ar(AEBC). Prove that DE || BC.
Ans: Given: D and E are points on sides AB and AC respectively of ∆ABC such that ar (∆DBC) = ar (∆EВС)
To Prove: DE || BC
Proof: ∵ ∆DBC and ∆EBC are on the same base BC and have equal areas.
∴ Their altitudes must be the same. ∴ 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).
Ans: Given: XY is a line parallel to side BC of a triangle ABC. BE || AC and CF || AB meets XY at E and F respectively.
To Prove: ar(ABE) = ar (ACF)
Proof: XY || BC | Given
and CF || BX | ∵ CF || AB (given)
| A quadrilateral is a parallelogram if the opposite sides are parallel
∴ BC = XF | Opposite sides of a parallelogram are equal
⇒ BC = XY+YF …(1)
Again,
∵ XY || BC | Given
and BE || CY | ∵ BE|| AC (given)
A quadrilateral is a parallelogram if the opposite sides are parallel
∴ BC = YE | Opposite sides of a parallelogram are equal
⇒ BC = XY + XE …(2)
From (1) and (2),
XY + YF = XY + XE
⇒ YF = XE
⇒ XE = YF …(3)
∵ ΔΑΕΧ and ΔAYF have equal bases (∵ XE = YF) on the same line EF and have a common vertex A.
∴ Their altitudes are also the same.
∴ ar (∆AEX) = ar (∆AFY) …(4)
∵ ∆BEX and ∆CFY have equal bases (∵ XE =YF) on the same line EF and are between the same parallels EF and BC (∵ XY || BC).
∴ ar (∆BEX) = ar (∆CFY) …(5) | Two triangles on the same base (or equal bases) and between the same parallels in area
Adding the corresponding sides of (4) and (5), we get
ar (∆AEX) + ar (∆BEX) = ar (∆AFY) + ar (∆CFY)
⇒ ar (∆ABE) = ar (∆ACF)
9. The side AB of a parallelo-gram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then par-allelogram PBQR is completed. Show that ar (∆ABCD) = ar(∆PBQR)
[Hint. Join AC and PQ. Now compare ar (ACQ) and ar (APQ)]
Ans: Given: The side AB of a paral-lelogram 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.
To Prove: ar (|| gm ABCD) = ar (|| gm PBQR)
Construction: Join AC and PQ.
Proof: ∵ AC is a diagonal of || gm ABCD
∵ ∆ACQ and ∆APQ are on the same base AQ and between the same parallels AQ and CP.
∴ ar (∆ACQ) = ar (∆APQ)
∵ Two triangles on the same base (or equal bases) and between the same parallels are equal in area
⇒ ar (∆ACQ)-ar (∆ABQ) = ar (∆APQ) – ur (∆ABQ) | Subtracting the same areas from both sides
⇒ ar (∆ABC) = ar (∆BPO)
⇒ ar (|| gm ABCD) = ar (|| gm PBQR).
10. Diagonals AC and BD of a trapezium ABCD with AB || DC in-tersect each other at O. Prove that ar (∆AOD) = ar (∆BOс).
Ans: Given: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.
To Prove: ar (∆AOD) = ar (∆BOC)
Το Proof: ∵ ΔABD and ∆ABC are on the same base AB and between the same parallels AB and DC
∴ ar (∆ABD) = ar (∆ABC)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area
⇒ ar (∆ABD) – ar (∆AOB) = ar (∆ABC) – ar (∆AOB) | Subtracting the same areas from both sides
⇒ ar (∆AOD) = ar (∆BOC)
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)
Ans: Given: ABCDE is a penta- gon. A line through B parallel to AC meets DC produced at F.
To Prove: (i) ar (∆ACB) = ar (ACF)
Proof: (i) ∵ ∆ACB and ∆ACF are on the same base AC and between the same parallels AC and BF.
[AC || BF (given)]
∴ ar (∆ACB) = ar (∆ACF)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area
(ii) From (i), ar (∆ACB) = ar (∆ACF)
12. A villager Itwarri has a plot of land in the shape of a quadrilat-eral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the concerns to construct a Health Centre. Itwari agrees to the above proposal with the condition that he should be given an 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.
Ans: Let ABCD be the plot of land of the shape of a quadrilateral. Let the portion ADE be taken over by the Gram Panchayat of the village from one corner D to construct a Health Centre.
Join AC. Draw a line through D par-allel to AC to meet BC produced in P. Then Itwaari must be given the land ECP adjoining his plot so as to form a triangular plot ABP as then.
ar (∆ADE) = ar (∆PEC)
Proof: ∵ ADPA and ADCP are on the same base DP and between the same parallels DP and AC.
∴ ar (∆ADE) = ar (∆DCP)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area
⇒ ar(∆DPA) – ar (∆DEP) = ar(∆DCP) – ar (∆DEP) | Subtracting the same areas from both sides
⇒ ar(∆ADE) = ar(∆PCE)
13. ABCD is a trapezium with AB || DC. A line parallel to AC inter-sects AB at X and BC at Y. Prove that ar (∆ADX) = ar (∆ACY).
[Hint. Join CX.]
Ans: Given: ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y.
To Prove: ar (∆ADX) = ar (∆ACY)
Construction: Join CX
Proof: ∵ ∆ ADX and ∆ACX are on the same base AX and between the same parallels AB and DC.
∴ ar (∆ADX) = ar(∆ACX) …(1)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area
∵ ∆ACX and ∆ACY are on the same base AC and between the same parallels AC and XY.
∴ ar(∆ACX) = ar(∆ACY) …(2)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area
From (1) and (2), we get, ar(∆ADX) = ar(∆ACY)
14. In figure, AP || BQ || CR. Prove that ar(∆ACQ) = ar(∆PBR)
Ans: Given: AP || BQ|| CR
To Prove: ar(∆ACQ) = ar(∆PBR)
Proof: ∆BAQ and ∆BPQ are on the same base BQ and between the same parallels BQ and AP
∴ ar(∆BAQ) = ar (∆BPQ) …(1)
| ∵ Two triangles on the same base (or equal bases) and between the same parallels are equal in area
∵ ∆BCQ and ∆BQR are on the same base BQ and between the same parallels BQ and CR.
∴ ar(∆BCQ) = ar (∆BQR) …(2)
| ∵ Two triangles on the same base (or equal bases) and Between the same parallels are equal in area
Adding the corresponding sides of (1) and (2), we get
ar(∆BAQ) = ar (∆BCQ) = ar (∆BPQ) + ar (∆BQR)
⇒ ar (∆AQC) =ar (∆PBR)
15. Diagonals AC and BD of a quadrilateral ABCD interested at O in such a way that ar(AOD) = ar(BOC). Prove that ABCD is a trapezium.
Ans: Given: Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (∆AOD) = ar (∆BOC)
Proof: ar (∆AOD) = ar (∆BOC)
⇒ ar (∆AOD) + ar (∆AOB) = ar (∆BOC) + ar (∆AOB)
| Adding the same areas on both sides
⇒ ar (∆ABD) = ar (∆ABC)
But ∆ABD and ∆ABC are on the same base AB.
∴ ∆ABD and ∆ABC will have equal corresponding altitudes.
∴ ∆ABC and ∆ABC will lie between the same parallels
∴ AB || DC
A quadrilateral is a trapezium if exactly one pair of opposite sides is parallel.
16. In figure, ar(∆DCR) = ar(∆DPC) and ar(∆BDP) = ar(∆ARC). Show that both the quadrilateral ABCD and DCPR are trapeziums.
Ans: Given: ar(∆DPC) = ar(∆DPC) and ar(∆BDP) = ar(∆ARC)
To Prove: Both the quadrilaterals ABCD and DCPR are trapeziums.
Proof: ar (∆DRC) = ar (∆DPC) …(1) | Given
But ∆DRC and ∆DPC are on the same base DC.
∴ ∆DRC and ∆DPC will have equal corresponding altitudes.
∴ ∆DRC and ∆DPC will lie between the same parallels.
∴ DC|| RP
| A quadrilateral is a trapezium if exactly one pair of opposite sides is parallel
Again, ar(∆BDP) = ar(∆ARC)
⇒ ar(∆BDC) + ar(∆DPC) = ar(∆ADC) + ar(∆DRC)
⇒ ar(∆BDC) = ar(∆ADC) | Using (1)
But ∆ BDC and ∆ ADC on the same base DC.
∴ ∆ BDC and ∆ ADC will have equal corresponding altitudes.
∴ ∆ BDC and ∆ ADC will lie between the same parallels.
∴ AB || DC
| A quadrilateral is a trapezium if exactly one pair of opposite sides is parallel.
