CLASS 9 MATHS CHAPTER-9 AREAS OF PARALLELOGRAMS AND TRIANGLES

Exercise 9.1

Question 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.

Solution:
In figure (i): DPC and trap. ABCD are on the same base DC and between same parallel DC and AB.

In figure (iii): RTQ and parallelogram PQRS are on the same base QR and between same parallel QR and PS.

In figure (v): Parallelogram ABCD and parallelogram APQD are on the same base AD and between the same parallels AD and BQ.

NCERT Solutions for Class 9 Maths Exercise 9.2

Question 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.

Solution:
BSOWe have, AE ⊥ DC and AB = 16 cm
∵ AB = CD [Opposite sides of parallelogram]
∴ CD = 16 cm
Now, area of parallelogram ABCD = CD x AE
= (16 x 8) cm2 = 128 cm2 [∵ AE = 8 cm]
Since, CF ⊥ AD
∴ Area of parallelogram ABCD = AD x CF
⇒ AD x CF = 128 cm
⇒ AD x 10 cm = 128 cm2 [∵ CF= 10 cm]
⇒ AD = cm = 12.8 cm 10
Thus, the required length of AD is 12.8 cm

Question 2. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = ar (ABCD).
Solution:
Join GE and HE, where GE || BC || DA and HF || AB || DC
(∵ E, F, G and H are the mid¬points of the sides of a ||gm ABCD).
If a triangle and a parallelogram are on the same base and between the same parallels, then A E U the area of the triangle is equal to half the area of the parallelogram.

Now, ∆EFG and parallelogram EBCG are on the same base EG and between the same parallels EG and BC.
∴ ar(∆EFG) = … (1)
Similarly, ar(∆EHG) = …(2)
Adding (1) and (2), we get
ar(∆EFG) + ar(∆EHG) = 
= 
Thus, ar(EFGH) = 

Question 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).
Solution:
∵ ABCD is a parallelogram.
∴ AB || CD and BC || AD.
Now, ∆APB and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.
∴ ar(∆APB) = …….(1)

Also, ∆BQC and parallelogram ABCD are on the same base BC and between the same parallels BGand AD.
∴ ar(∆BQC) = …(2)
From (1) and (2), we have ar(∆APB) = ar(∆BQC).

Question 4. In figure, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar (APB) + ar (PCD) = 
(ii) ar (APD) + ar(PBC) = ar (APB) + ar (PCD)
Solution:
We have a parallelogram ABCD, i.e., AB || CD and BC || AD. Let us draw EF || AB and HG || AD through P.

(i) ∆APB and ||gm AEFB are on the same base AB and between the same parallels AB and EF.
∴ ar(∆APB) = …(1)
Also, ∆PCD and parallelogram CDEF are on the same base CD and between the same parallels CD and EF.
∴ ar(APCD) = …(2)
Adding (1) and (2), we have
ar(∆APB) + ar(∆PCD) = 
⇒ ar(∆APB) + ar(∆PCD) = …(3)

(ii) ∆APD and ||gm ∆DGH are on the same base AD and between the same parallels AD and GH.
∴ ar(∆APD) = …(4)
Similarly,
ar(∆PBC) = …(5)
Adding (4) and (5), we have
ar(∆APD) + ar(∆PBC) = = 
⇒ ar(∆APD) + ar(∆PBC) = …….(6)
From (3) and (6), we have
ar(∆APD) + ar(∆PBC) = ar(∆APB) + ar(∆PCD)

Question 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) = 

Solution:
(i) Parallelogram PQRS and parallelogram ABRS are on the same base RS and between the same parallels RS and PB.
∴ ar(PQRS) = ar(ABRS)
(ii) AAXS and ||gm ABRS are on the same base AS and between the same parallels AS and BR. *
∴ ar(AXS) = …(1)
But ar(PQRS) = ar(ABRS) …(2) [Proved in (i) part]
From (1) and (2), we have
ar(AXS) = 

Question 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.
Solution:
The farmer is having the field in the form of parallelogram PQRS and a point A is situated on RS. Join AP and AQ.
Clearly, the field is divided into three parts i.e., in ∆APS, ∆PAQ and ∆QAR.

Since, ∆PAQ and pt.
parallelogram PQRS are on the same base PQ and between the same parallels PQ and RS.
ar(∆PAQ) = …(1)
⇒ ar(||gm PQRS) – ar(∆PAQ) = ar(||gm PQRS) – 
⇒ [ar(∆APS) + ar(∆QAR)] = …(2)
From (1) and (2), we have
ar(∆PAQ) = ar[(∆APS) + (∆QAR)]
Thus, the farmer can sow wheat in (∆PAQ) and pulses in [(∆APS) + (∆QAR)] or wheat in [(∆APS) + (∆QAR)] and pulses in (∆PAQ).

NCERT Solutions for Class 9 Maths Exercise 9.3

Question 1. In figure, E is any point on median AD of a ABC. Show that ar (ABE) = ar (ACE).

Solution:
In ABC, AD is a median.

ar (ABD) = ar (ACD) ……….(i)

[Median divides a into two s of equal area]

Again in EBC, ED is a median

ar (EBD) = ar (ECD) ……….(ii)

Subtracting eq. (ii) from (i),

ar (ABD) – ar (EBD) = ar (ACD) – ar (ECD)

ar (ABE) = ar (ACE)

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

Solution: 
Given: A ABC, AD is the median and E is the mid-point of median AD.

To prove: ar (BED) = ar (ABC)

Proof: In ABC, AD is the median.

ar (ABD) = ar (ADC)

[Median divides a into two s of equal area]

ar (ABD) = ar (ABC) ……….(i)

In ABD, BE is the median.

ar (BED) = ar (BAE)

ar (BED) = ar (ABD)

ar (BED) = ar (ABC) = ar (ABC)

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

Solution: 
Let parallelogram be ABCD and its diagonals AC and BD intersect each other at O.

In ABC and ADC,

AB = DC [Opposite sides of a parallelogram]

BC = AD [Opposite sides of a parallelogram]

And AC = AC [Common]

ABC CDA [By SSS congruency]

Since, diagonals of a parallelogram bisect each other.

O is the mid-point of bisection.

Now in ADC, DO is the median.

ar (AOD) = ar (COD) ……….(i)

[Median divides a triangle into two equal areas]

Similarly, in ABC, OB is the median.

ar (AOB) = ar (BOC) ……….(ii)

And in AOB and AOD, AO is the median.

ar (AOB) = ar (AOD) ……….(iii)

From eq. (i), (ii) and (iii),

ar (AOB) = ar (AOD) = ar (BOC) = ar (COD)

Thus diagonals of parallelogram divide it into four triangles of equal area.

Question 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).

Solution: 
Draw CMAB and DNAB.

In CMO and DNO,

CMO = DNO = [By construction]

COM = DON [Vertically opposite]

OC = OD [Given]

CMO DNO [By ASA congruency]

AM = DN [By CPCT] ……(i)

Now ar (ABC) = ……….(ii)

ar (ADB) = ……….(iii)

Using eq. (i) and (iii),

ar (ADB) = ……….(iv)

From eq. (ii) and (iv),

ar (ABC) = ar (ADB)

Question 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.

(ii) ar (DEF) = ar (ABC)

(iii) ar (BDEF) = ar (ABC)

Solution: 
(i) F is the mid-point of AB and E is the mid-point of AC.

FEBC and FE = BD

[ Line joining the mid-points of two sides of a triangle is parallel to the third and half of it]

FEBD [BD is the part of BC]

And FE = BD

Also, D is the mid-point of BC.

BD = BC

And FEBC and FE = BD

Again E is the mid-point of AC and D is the mid-point of BC.

DEAB and DE = AB

DEAB [BF is the part of AB]

And DE = BF

Again F is the mid-point of AB.

BF = AB

But DE = AB

DE = BF

Now we have FEBD and DEBF

And FE = BD and DE = BF

Therefore, BDEF is a parallelogram.

(ii) BDEF is a parallelogram.

ar (BDF) = ar (DEF) ……….(i)

[diagonals of parallelogram divides it in two triangles of equal area]

DCEF is also parallelogram.

ar (DEF) = ar (DEC) ……….(ii)

Also, AEDF is also parallelogram.

ar (AFE) = ar (DEF) ……….(iii)

From eq. (i), (ii) and (iii),

ar (DEF) = ar (BDF) = ar (DEC) = ar (AFE) ……….(iv)

Now, ar (ABC) = ar (DEF) + ar (BDF) + ar (DEC) + ar (AFE) ……….(v)

ar (ABC) = ar (DEF) + ar (DEF) + ar (DEF) + ar (DEF)

[Using (iv) & (v)]

ar (ABC) = 4 ar (DEF)

ar (DEF) = ar (ABC)

(iii) ar (gm BDEF) = ar (BDF) + ar (DEF) = ar (DEF) + ar (DEF) [Using (iv)]

ar (gm BDEF) = 2 ar (DEF)

ar (gm BDEF) = ar (ABC)

ar (gm BDEF) = ar (ABC)

Question 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) DACB or ABCD is a parallelogram.

Solution: 
(i) Draw BM AC and DN AC.

In DON and BOM,

OD = OB [Given]

DNO = BMO = [By construction]

DON = BOM [Vertically opposite]

DON BOM [By RHS congruency]

DN = BM [By CPCT]

Also ar (DON) = ar (BOM) ……….(i)

Again, In DCN and ABM,

CD = AB [Given]

DNC = BMA = [By construction]

DN = BM [Prove above]

DCN BAM [By RHS congruency]

ar (DCN) = ar (BAM) ……….(ii)

Adding eq. (i) and (ii),

ar (DON) + ar (DCN) = ar (BOM) + ar (BAM)

ar (DOC) = ar (AOB)

(ii) Since ar (DOC) = ar (AOB)

Adding ar BOC both sides,

ar (DOC) + ar BOC = ar (AOB) + ar BOC

ar (DCB) = ar (ACB)

(iii) Since ar (DCB) = ar (ACB)

Therefore, these two triangles in addition to be on the same base CB lie between two same parallels CB and DA.

DACB

Now AB = CD and DACB

Therefore, ABCD is a parallelogram.

Question 7. D and E are points on sides AB and AC respectively of ABC such that ar (DBC) = ar (EBC). Prove that DEBC.

Solution:
Given: ar (DBC) = ar (EBC)

Since two triangles of equal area have common base BC.

Therefore, DEBC [Two triangles having same base (or equal bases) and equal areas lie between the same parallel]

Question 8. XY is a line parallel to side BC of triangle ABC. If BEAC and CFAB meet XY at E and F respectively, show that ar (ABE) = ar (ACF).

Solution:
ABE and parallelogram BCYE lie on the same base BE and between the same parallels BE and AC.

ar (ABE) = ar (gm BCYE) ……….(i)

Also ACF and gm BCFX lie on the same base CF and between same parallel BX and CF.

ar (ACF) = ar (gm BCFX) ……….(ii)

But gm BCYE and gm BCFX lie on the same base BC and between the same parallels BC and EF.

ar (gm BCYE) = ar (gm BCFX) ……….(iii)

From eq. (i), (ii) and (iii), we get,

ar (ABE) = ar (ACF)

Question 9. The side AB of 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. Show that ar (ABCD) = ar (PBQR).

Solution: 
Given: ABCD is a parallelogram, CPAQ and PBQR is a parallelogram.

To prove: ar (ABCD) = ar (PBQR)

Construction: Join AC and QP.

Proof: Since AQCP

ar (AQC) = ar (AQP)

[Triangles on the same base and between the same parallels are equal in area]

Subtracting ar (ABQ) from both sides, we get

ar (AQC – ar (ABQ) = ar (AQP) – ar (ABQ)

ar (ABC) = ar (QBP) ……….(i)

Now ar (ABC) = ar (gm ABCD)

[Diagonal divides a parallelogram in two parts of equal area]

And ar (PQB) = ar (gm PBQR)

From eq. (i), (ii) and (iii), we get

ar (gm ABCD) = ar (gm PBQR)

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

Solution:
ABD and ABC lie on the same base AB and between the same parallels AB and DC.

ar (ABD) = ar (ABC)

Subtracting ar (AOB) from both sides,

ar (ABD) – ar (AOB)

= ar (ABC) – ar (AOB)

ar (AOD) = ar (BOC)

Question 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)

Solution: 
(i) Given that BFAC

ACB and ACF lie on the same base AC and between the same parallels AC and BF.

ar (ACB) = ar (ACF) ……….(i)

(ii) Now ar (ABCDE) = ar (trap. AEDC) + ar (ABC) ……….(ii)

ar (ABCDE) = ar (trap. AEDC) + ar (ACF) = ar (quad. AEDF) [Using (i)]

ar (AEDF) = ar (ABCDE)

Question 12. A villager Itwaari has a plot of land of the shape of quadrilateral. The Gram Panchyat of two villages decided to take over some portion of his plot from one of the corners to construct a health centre. Itwaari agrees to the above personal 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.

Solution:
Let Itwari has land in shape of quadrilateral PQRS.

Draw a line through 5 parallel to PR, which meets QR produced at M.

Let diagonals PM and RS of new formed quadrilateral intersect each other at point N.

We have PRSM [By construction]

ar (PRS) = ar (PMR)

[Triangles on the same base and same parallel are equal in area]

Subtracting ar (PNR) from both sides,

ar (PRS) – ar (PNR) = ar (PMR) – ar (PNR)

ar (PSN) = ar (MNR)

It implies that Itwari will give corner triangular shaped plot PSN to the Grampanchayat for health centre and will take equal amount of land (denoted by MNR) adjoining his plot so as to form a triangular plot PQM.

Question 13. ABCD is a trapezium with ABDC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

Solution:
Join CX, ADX and ACX lie on the same base XA and between the same parallels XA and DC.

ar (ADX) = ar (ACX) ……….(i)

Also ACX and ACY lie on the same base

AC and between same parallels CY and XA.

ar (ACX) = ar (ACY) ……….(ii)

From (i) and (ii),

ar (ADX) = ar (ACY)

Question 14. In figure, APBQCR. Prove that ar (AQC) = ar (PBR).

Solution:
ABQ and BPQ lie on the same base BQ and between same parallels AP and BQ.

ar (ABQ) = ar (BPQ) ……….(i)

BQC and BQR lie on the same base BQ and between same parallels BQ and CR.

ar (BQC) = ar (BQR) ……….(ii)

Adding eq (i) and (ii), ar (ABQ) + ar (BQC) = ar (BPQ) + ar (BQR)

ar (AQC) = ar (PBR)

Question 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.

Solution:
Given that ar (AOD) = ar (BOC)

Adding AOB both sides,

ar (AOD) + ar (AOB) = ar (BOC) + ar (AOB)

ar (ABD) = ar (ABC)

Since if two triangles equal in area, lie on the same base then, they lie between same parallels. We have ABD and ABC lie on common base AB and are equal in area.

They lie in same parallels AB and DC.

ABDC

Now in quadrilateral ABCD, we have ABDC

Therefore, ABCD is trapezium. [In trapezium one pair of opposite sides is parallel]

Question 16. In figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution:
Given that DRC and DPC lie on the same base DC and ar (DPC) = ar (DRC) …..(i)

DCRP

[If two triangles equal in area, lie on the same base then, they lie between same parallels]

Therefore, DCPR is trapezium. [ In trapezium one pair of opposite sides is parallel]

Also ar (BDP) = ar (ARC) ……….(ii)

Subtracting eq. (i) from (ii),

ar (BDP) – ar (DPC) = ar (ARC) – ar (DRC)

ar (BDC) = ar (ADC)

Therefore, ABDC [If two triangles equal in area, lie on the same base then, they lie between same parallels]

Therefore, ABCD is trapezium.

NCERT Solutions for Class 9 Maths Exercise 9.4

Question 1. Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Solution:
We have a parallelogram ABCD and rectangle ABEF such that
ar(||gm ABCD) = ar( rect. ABEF)

AB = CD [Opposite sides of parallelogram]
and AB = EF [Opposite sides of a rectangle]
⇒ CD = EF
⇒ AB + CD = AB + EF … (1)
BE < BC and AF < AD [In a right triangle, hypotenuse is the longest side] ⇒ (BC + AD) > (BE + AF) …(2)
From (1) and (2), we have
(AB + CD) + (BC+AD) > (AB + EF) + BE + AF)
⇒ (AB + BC + CD + DA) > (AB + BE + EF + FA)
⇒ Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF.

Question 2. In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).

Solution:
Let us draw AF, perpendicular to BC
such that AF is the height of ∆ABD, ∆ADE and ∆AEC.

Question 3. In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ax(BCF).

Solution:
Since, ABCD is a parallelogram [Given]
∴ Its opposite sides are parallel and equal.
i.e., AD = BC …(1)
Now, ∆ADE and ∆BCF are on equal bases AD = BC [from (1)] and between the same parallels AB and EF.
So, ar(∆ADE) = ar(∆BCF).

Question 4. In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar(BPC) = ax(DPQ).[Hint Join AC.]

Solution:
We have a parallelogram ABCD and AD = CQ. Let us join AC.
We know that triangles on the same base and between the same parallels are equal in area.
Since, ∆QAC and ∆QDC are on the same base QC and between the same parallels AD and BQ.
∴ ar(∆QAC) = ar(∆QDC)
Subtracting ar(∆QPC) from both sides, we have
ar(∆QAQ – ar(∆QPC) = ar(∆QDC) – ar(∆QPC)
⇒ ar(∆PAQ = ar(∆QDP) …(1)
Since, ∆PAC and ∆PBC are on the same base PC and between the same parallels AB and CD.
∴ ar(∆PAC) = ar(∆PBC) …(2)
From (1) and (2), we get
ar(∆PBC) = ar(∆QDP)

Question 5. In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, Show that

[Hint Join EC and AD. Show that BE || AC and DE || AB, etc.]

Solution:
Let us join EC and AD. Draw EP ⊥ BC.
Let AB = BC = CA = a, then
BD = = DE = BE

(ii) Since, ∆ABC and ∆BED are equilateral triangles.
⇒ ∠ACB = ∠DBE = 60°
⇒ BE || AC
∆BAE and ∆BEC are on the same base BE and between the same parallels BE and AC.
ar(∆BAE) = ar(∆BEC)
⇒ ar(∆BAE) = 2 ar(∆BDE) [ DE is median of ∆EBC. ∴ ar(∆BEC) = || ar(∆BDE)]
⇒ ar(ABDE) = ar(∆BAE)

(iii) ar(∆ABC) = 4 ar(∆BDE)[Proved in (i) part]
ar(∆BEC) = 2 ar(∆BDE)
[ ∵ DE is median of ∆BEC]
⇒ ar(∆ABC) = 2 ar(∆BEC)

(iv) Since, ∆ABC and ∆BDE are equilateral triangles.
⇒ ∠ABC = ∠BDE = 60°
⇒ AB || DE
∆BED and ∆AED are on the same base ED and between the same parallels AB and DE.
∴ ar(∆BED) = ar(∆AED)
Subtracting ar(AEFD) from both sides, we get
⇒ ar(∆BED) – ar(∆EFD) = ar(∆AED) – ar(∆EFD)
⇒ ar(∆BEE) = ar(∆AFD)

(v) In right angled ∆ABD, we get

From (1) and (2), we get
ar(∆AFD) = 2 ar(∆EFD)
ar(∆AFD) = ar(∆BEF) [From (iv) part]
⇒ ar(∆BFE) = 2 ar(∆EFD)

(vi) ar(∆AFC) = ar(∆AFD) + ar(∆ADC)
= ar(∆BFE) + ar(∆ABC) [From (iv) part]
= ar(∆BFE) + x 4 x ar(∆BDE) [From (i) part]
= ar(∆BFE) + 2ar(∆BDE)
= 2ar(∆FED) + 2[ar(∆BFE) + ar(∆FED)]
= 2ar(∆FED) + 2[2ar(∆FED) + ar(∆FED)] [From (v) part]
= 2ar(∆FED) + 2[3ar(∆FED)]
= 2ar(∆FED) + 6ar(∆FED)
= 8ar(∆FED)
∴ ar(∆FED) = ar(∆AFC)

Question 6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar(APB) x ar(CPD) = ar(APD) x ar(BPC).
[Hint From A and C, draw perpendiculars to BD.]

Solution:
We have a quadrilateral ABCD such that its diagonals AC and BD intersect at P.
Let us draw AM ⊥ BD and CN ⊥ BD.

Question 7. P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that

Solution:
We have a ∆ABC such that P is the mid-point of AB and Q is the mid-point of BC.
Also, R is the mid-point of AP. Let us join AQ, RQ, PC and PC.

(i) In ∆APQ, R is the mid-point of AP. [Given] B

∴RQ is a median of ∆APQ.
⇒ ar(∆PRQ) = ar(∆APQ) …(1)
In ∆ABQ, P is the mid-point of AB.
∴ QP is a median of ∆ABQ.
∴ ar(∆APQ) = ar(∆ABQ) …(2)

Question 8. In figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that

(i) ∆MBC = ∆ABD
(ii) ar(BYXD) = 2 ar(MBC)
(iii) ar(BYXD) = ax(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar(CYXE) = ax(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)

Solution:
We have a right ∆ABC such that BCED, ACFG and ABMN are squares on its sides BC, CA and AB respectively. Line segment AX 1 DE is also drawn such that it meets BC at Y.

(i) ∠CBD = ∠MBA [Each90°]
∴ ∠CBD + ∠ABC = ∠MBA + ∠ABC
(By adding ∠ABC on both sides)
or ∠ABD = ∠MBC
In ∆ABD and ∆MBC, we have
AB = MB [Sides of a square]
BD = BC
∠ABD = ∠MBC [Proved above]
∴ ∆ABD = ∆MBC [By SAS congruency]

(ii) Since parallelogram BYXD and ∆ABD are on the same base BD and between the same parallels BD and AX.
∴ ar(∆ABD) = ar(||gm BYXD)
But ∆ABD ≅ ∆MBC [From (i) part]
Since, congruent triangles have equal
areas.
∴ ar(∆MBC) = ar(||gm BYXD)
⇒ ar(||gm BYXD) = 2ar(∆MBC)

(iii) Since, ar(||gm BYXD) = 2ar(∆MBC) …(1) [From (ii) part]
and or(square ABMN) = 2or(∆MBC) …(2)
[ABMN and AMBC are on the same base MB and between the same parallels MB and NC]
From (1) and (2), we have
ar(BYXD) = ar(ABMN) .

(iv) ∠FCA = ∠BCE (Each 90°)
or ∠FCA+ ∠ACB = ∠BCE+ ∠ACB
[By adding ∠ACB on both sides]
⇒ ∠FCB = ∠ACE
In ∆FCB and ∆ACE, we have
FC = AC [Sides of a square]
CB = CE [Sides of a square]
∠FCB = ∠ACE [Proved above]
⇒ ∆FCB ≅ ∆ACE [By SAS congruency]

(v) Since, ||gm CYXE and ∆ACE are on the same base CE and between the same parallels CE and AX.
∴ ar(||gm CYXE) = 2ar(∆ACE)
But ∆ACE ≅ ∆FCB [From (iv) part]
Since, congruent triangles are equal in areas.
∴ ar (||<gm CYXE) = 2ar(∆FCB)

(vi) Since, ar(||gm CYXE) = 2ar(∆FCB) …(3)
[From (v) part]
Also (quad. ACFG) and ∆FCB are on the same base FC and between the same parallels FC and BG.
⇒ ar(quad. ACFG) = 2ar(∆FCB) …(4)
From (3) and (4), we get
ar(quad. CYXE) = ar(quad. ACFG) …(5)

(vii) We have ar(quad. BCED)
= ar(quad. CYXE) + ar(quad. BYXD)
= ar(quad. CYXE) + ar(quad. ABMN)
[From (iii) part]
Thus, ar (quad. BCED)
= ar(quad. ABMN) + ar(quad. ACFG)
[From (vi) part]