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Page No 5:

Question 1:

Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.

Answer:

Let ABC and PQR be two right triangles with AB ⊥ BC and PQ ⊥ QR.
Given:
BC = 9, AB = 5, PQ = 6 and QR = 10.
AABCAPQR=AB×BCPQ×QR=5×96×10=34



Page No 6:

Question 2:

In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find AABCAADB.

Answer:

Given:
BC = 4
AD = 8
AABCAADB=AB×BCAB×AD=BCAD                             BC=4 and AD=8=48
=12

Page No 6:

Question 3:

In adjoining figure, seg PS ⊥ seg RQ seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12, then Find QT.

Answer:

Given:
PS ⊥ RQ
QT ⊥ PR
RQ = 6, PS = 6 and PR = 12
With base PR and height QT, APQR=12×PR×QT
With base QR and height PS, APQR=12×QR×PS
APQRAPQR=12×PR×QT12×QR×PS1=PR×QTQR×PSPR×QT=QR×PS
QT=QR×PSPR=6×612=3
Hence, the measure of side QT is 3 units.

Page No 6:

Question 4:

In adjoining figure, AP ⊥ BC, AD || BC, then Find A(∆ABC) : A(∆BCD).

Answer:

Given:
AP ⊥ BC
AD || BC
AABCABCD=AP×BCAP×BC=11
Hence, the ratio of A(∆ABC) and A(∆BCD) is 1 : 1.
 

Page No 6:

Question 5:

In adjoining figure PQ ⊥ BC, AD⊥ BC then find following ratios.


(i) APQBAPBC

(ii) APBCAABC

(iii) AABCAADC

(iv) AADCAPQC

 

Answer:

(i)
APQBAPBC=PQ×BQPQ×BC=BQBC
(ii)
APBCAABC=PQ×BCAD×BC=PQAD
(iii)
AABCAADC=AD×BCAD×DC=BCDC
(iv)
AADCAPQC=AD×DCPQ×QC



Page No 13:

Question 1:

Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR.
(1)

(2)

(3)

Answer:

(1)
In QMP, QMQP=3.57=12
In MRP,MRRP=1.53=12
QMQP=MRRP
By converse of angle bisector theorem, ray PM is the bisector of ∠QPR.

(2)
In QMP, QMQP=810=45
In MRP,MRRP=67
QMQPMRRP
Therefore, ray PM is not the the bisector of ∠QPR.
(3)
In QMP, QMQP=3.69=25
In MRP,MRRP=410=25
QMQP=MRRP
By converse of angle bisector theorem, ray PM is the bisector of ∠QPR.


 

Page No 13:

Question 2:

In ∆PQR, PM = 15, PQ = 25 PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.

Answer:

Given:
PM = 15,
PQ = 25,
PR = 20 and
NR = 8
Now, PN = PR − NR
= 20 − 8
= 12
Also, MQ = PQ − PM
= 25 − 15
= 10
In PRQ, PRNR=128=32
Also,PMMQ=1510=32
PRNR=PMMQ
By converse of basic proportionality theorem, NM is parallel to side RQ or NM || RQ.



Page No 14:

Question 3:

In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then Find QP.

Answer:

In PNM, QMQP=MNPN                      By angle bisector theorem2.5QP=57
QP=2.5×75=3.5
Hence, the measure of QP is 3.5.

Page No 14:

Question 4:

Measures of some angles in the figure are given. Prove that APPB=AQQC

Answer:

Given:
∠APQ = 60
∠ABC = 60
Since, the corresponding angles ∠APQ and ​∠APC are equal.
Hence, line PQ || BC.
In ABC, PQBCAPPB=AQQC                      By Basic proportionalitytheorem

Page No 14:

Question 5:

In trapezium ABCD, side AB || side PQ || side ∆C, AP = 15, PD = 12, QC = 14, Find BQ.

Answer:

Construction: Join BD intersecting PQ at X.

In △ABD, PX || AB
DPPA=DXXB                   ...1     By Basic proportionality theorem
In △BDC, XQ||DC 
DXXB=CQQB                   ...2     By Basic proportionality theorem
From (1) and (2), we get
DPPA=CQQB1215=14QBQB=17.5

Page No 14:

Question 6:

Find QP using given information in the figure.

Answer:

In PNM, QMQP=MNPN                      By angle bisector theorem14QP=2540
QP=14×4025=22.4
Hence, the measure of QP is 22.4.

Page No 14:

Question 7:

In the given figure, if AB || CD || FE then Find x and AE.

Answer:

Construction: Join AFintersecting CD at X.

In △ABF, DX || AB
FDDB=FXXA               ...1           By Basic proportionality theorem
In △AEF, XC||FE 
FXXA=ECCA               ...2            By Basic proportionality theorem
From (1) and (2), we get
FDDB=ECCA48=x12x=6
Now, AE = AC + CE
= 12 + 6
= 18



Page No 15:

Question 8:

In ∆LMN, ray MT bisects ∠LMN If LM = 6, MN = 10, TN = 8, then Find LT.

Answer:

In LNM, LTNT=LMNM                      By angle bisector theoremLT8=610
LT=8×610=4.8
Hence, the measure of LT is 4.8.

Page No 15:

Question 9:

In ∆ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.

Answer:

In △ABC, ∠ABD = ∠DBC
ADDC=ABCB                            By angle bisector theoremx-2x+2=xx+5x2+3x-10=x2+2x
3x-2x=10x=10

Page No 15:

Question 10:

In the given figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.

Answer:

Given:
Seg PQ || seg DE
seg QR || seg EF
In △DXE, PQ || DE
XPPD=XQQE                           ...I By basic proportionality theorem
In △XEF, QR || EF                ....Given
XQQE=XRRF                           .....II By basic proportionality theorem
XPPD=XRRF                          From I and II
∴ seg PR || seg DF        (Converse of basic proportional theorem)

Page No 15:

Question 11:

In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC.

Answer:

Given:
ray BD bisects ∠ABC
ray CE bisects ∠ACB.
seg AB ≅ seg AC

In △ABC, ∠ABD = ∠DBC
ADDC=ABBC                           ...I By angle bisector theorem
In △ABC, ∠BCE = ∠ACE
AEEB=ACBC                           ...II By angle bisector theorem
From (I) and (II)
ADDC=AEEB             seg AB  seg AC
∴ ​ED || BC        (Converse of basic proportional theorem)



Page No 21:

Question 1:

In the given figure, ∠ABC = 75°, ∠EDC = 75° state which two triangles are similar and by which test? Also write the similarity of these two triangles by a proper one to one correspondence.

Answer:

Given: 
∠ABC = 75°, ∠EDC = 75°
Now, in △ABC and △EDC
∠ABC = ∠EDC = 75°         (Given)
∠C = ∠C         (Common)
By AA test of similarity
△ABC ∼ △EDC

Page No 21:

Question 2:

Are the triangles in the given figure similar? If yes, by which test ?

Answer:

Given: 
PQ = 6
PR = 10
QR = 8
LM = 3
LN = 5
MN = 4
Now,
 PQLM=63=2,QRMN=84=2,RPNL=105=2
PQLM=QRMN=RPNL
By SSS test of similarity
△PQR ∼ △LMN

Page No 21:

Question 3:

As shown in the given figure, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m then how long will be the shadow of the bigger pole at the same time ?

Answer:

Given: 
PR = 4
RL = 6
AC = 8
In △PLR and △ABC
∠PRL = ∠ACB         (Vertically opposite angles)
∠LPR = ∠BAC         (Angles made by sunlight on top are congruent)
By AA test of similarity
△PLR ∼ △ABC
PRAC=LRBC                      Corresponding sides are proportional48=6xx=12
Hence, the length of shadow of bigger pole due to sunlight is 12 m.

Page No 21:

Question 4:

In ∆ABC, AP ⊥ BC, BQ ⊥ AC B– P–C, A–Q – C then prove that, ∆CPA ~ ∆CQB. If AP = 7, BQ = 8, BC = 12 then Find AC.

Answer:

Given: 
AP ⊥ BC
BQ ⊥ AC
To prove: ∆CPA ~ ∆CQB
Proof: In ∆CPA and ∆CQB
∠CPA = ∠CQB = 90         (Given)
∠C = ∠C                             (Common)
By AA test of similarity
∆CPA ~ ∆CQB
Hence proved.
Now, APBQ=ACBC                      Corresponding sides are proportionalAC=APBQ×BC=78×12=10.5
 



Page No 22:

Question 5:

Given : In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ

Answer:

Given: 
side PQ || side SR
AR = 5AP,
AS = 5AQ
To prove: SR = 5PQ
Proof: In ∆APQ and ∆ARS
∠PAQ = ∠RAS          (Vertically Opposite angles)
∠PQA = ∠RSA          (Alternate angles, side PQ || side SR and QS is a transversal line)
By AA test of similarity
∆APQ ~ ∆ARS
PQSR=APAR                      Corresponding sides are proportionalPQSR=15                  AR=5APSR=5PQ
Hence proved.

Page No 22:

Question 6:

In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then Find OD.

Answer:

Given: 
side AB || side DC
AB = 20,
DC = 6,
OB = 15
In △COD and △AOB
∠COD = ∠AOB         (Vertically opposite angles)
∠CDO= ∠ABO         (Alternate angles, CD ||BA and BD is a transversal line)
By AA test of similarity
△COD ∼ △AOB
CDAB=ODOB                      Corresponding sides are proportional620=OD15OD=4.5

Page No 22:

Question 7:

◻ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE.

Answer:

Given: ◻ABCD is a parallelogram
To prove: DE × BE = CE × TE
Proof: In ∆BET and ∆CED
∠BET = ∠CED         (Vertically opposite angles)
∠BTE = ∠CDE         (Alternate angles, AT || CD and DT is a transversal line)
By AA test of similarity
∆BET ∼ ∆CED
BECE=ETED                      Corresponding sides are proportionalBE×ED=CE×ET
Hence proved.

Page No 22:

Question 8:

In the given figure, seg AC and seg BD intersect each other in point P and APCP=BPDP. Prove that, ∆ABP ~ ∆CDP

Answer:

Given: APCP=BPDP
To prove: ∆ABP ~ ∆CDP
Proof: In ∆ABP and ∆DCP
APCP=BPDP       (Given)
∠P = ∠P                   (Common)
By SAS test of similarity
APCP=BPDP

Page No 22:

Question 9:

In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA2 = CB × CD

Answer:

Given:  ∠BAC = ∠ADC
To prove: CA2 = CB × CD
Proof: In ∆ABC and ∆DAC
∠BAC = ∠ADC       (Given)
∠C = ∠C                   (Common)
By AA test of similarity
∆ABC ∼ ∆DAC
BCAC=ACDC                              Corresponding sides are proportionalAC2=BC×DC
Hence proved.



Page No 25:

Question 1:

The ratio of corresponding sides of similar triangles is 3 : 5; then Find the ratio of their areas.

Answer:

According to theorem of areas of similar triangles "When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides".
Therefore, the ratio of the areas of triangles =3252
=925

Page No 25:

Question 2:

If ∆ABC ~ ∆PQR and AB : PQ = 2 : 3, then fill in the blanks.
AABCAPQR=AB2       =2232=              

Answer:

Given:
∆ABC ~ ∆PQR
AB : PQ = 2 : 3
According to theorem of areas of similar triangles "When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides".
AABCAPQR=AB2PQ2=2232=    4       9   

Page No 25:

Question 3:

If ∆ABC ~ ∆PQR, A (∆ABC) = 80, A (∆PQR) = 125, then fill in the blanks.
AABCA. . . .=80125            ABPQ=                  

Answer:

Given:
∆ABC ~ ∆PQR
A (∆ABC) = 80
A (∆PQR) = 125
According to theorem of areas of similar triangles "When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides".
AABCAPQR=AB2PQ280125=AB2PQ21625=AB2PQ2
4252=AB2PQ2ABPQ=45
Therefore, AABCAPQR=80125 and ABPQ=45
 

Page No 25:

Question 4:

∆LMN ~ ∆PQR, 9 × A (∆PQR ) = 16 × A (∆LMN). If QR = 20 then Find MN.

Answer:

Given:
∆LMN ~ ∆PQR
9 × A (∆PQR ) = 16 × A (∆LMN)
Consider, 9 × A (∆PQR ) = 16 × A (∆LMN)
ALMNAPQR=916MN2QR2=3242MNQR=34
MN=34×QRMN=34×20                  QR=20MN=15
 

Page No 25:

Question 5:

Areas of two similar triangles are 225 sq.cm. 81 sq.cm. If a side of the smaller triangle is 12 cm, then Find corresponding side of the bigger triangle.

Answer:

According to theorem of areas of similar triangles "When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides".
Area of bigger triangleArea of smaller triangle=22581Side of bigger triangle2Side of smaller triangle2=15292Side of bigger triangleSide of smaller triangle=159
Side of bigger triangle=159×Side of smaller triangleSide of bigger triangle=159×12=20

Hence, the corresponding side of the bigger triangle is 20.

Page No 25:

Question 6:

∆ABC and ∆DEF are equilateral triangles. If A(∆ABC) : A (∆DEF) = 1 : 2 and AB = 4, find DE.

Answer:

Consider, A(∆ABC) : A (∆DEF) = 1 : 2
AABCADEF=12AB2DE2=12DE2=2AB2
DE2=2×42                  AB=4DE=32DE=42

Page No 25:

Question 7:

In the given figure 1.66, seg PQ || seg DE, A(∆PQF) = 20 units, PF = 2 DP, then Find A(◻DPQE) by completing the following activity.

Answer:

Given: 
seg PQ || seg DE
A(∆PQF) = 20 units
PF = 2 DP
Let us assume DP = x
∴ PF = 2x
DF=DP+PF=x+2x=3x
In △FDE and △FPQ
∠FDE = ∠FPQ         (Corresponding angles)
∠FED = ∠FQP         (Corresponding angles)
By AA test of similarity
△FDE ∼ △FPQ
AFDEAFPQ=FD2FP2=3x22x2=94AFDE=94AFPQ=94×20=45
ADPQE=AFDE-AFPQ=45-20=25

 



Page No 26:

Question 1:

Select the appropriate alternative.
(1) In ∆ABC and ∆PQR, in a one to one correspondence ABQR=BCPR=CAPQ then

(A) ∆PQR ~ ∆ABC
(B) ∆PQR ~ ∆CAB
(C) ∆CBA ~ ∆PQR
(D) ∆BCA ~ ∆PQR

(2) If in ∆DEF and ∆PQR, ∠D ≅ ∠Q, ∠R ≅ ∠E then which of the following statements is false?

(A) EFPR=DFPQ (B) DEPQ=EFRP

(C) DEQR=DFPQ (D) EFRP=DEQR



(3) In ∆ABC and ∆DEF ∠B = ∠E, ∠F = ∠C and AB = 3DE then which of the statements regarding the two triangles is true ?
(A)The triangles are not congruent and not similar
(B) The triangles are similar but not congruent.
(C) The triangles are congruent and similar.
(D) None of the statements above is true.


(4) ∆ABC and ∆DEF are equilateral triangles, A (∆ABC) : A (∆DEF) = 1 : 2
If AB = 4 then what is length of DE?
(A) 22
(B) 4
(C) 8
(D) 42



(5) In the given figure, seg XY || seg BC, then which of the following statements is true?

(A) ABAC=AXAY (B) AXXB=AYAC

(C) AXYC=AYXB (D) ABYC=ACXB

Answer:

(1)
Given: ABQR=BCPR=CAPQ
By SSS test of similarity
 ∆PQR ~ ∆CAB
Hence, the correct option is (B).

(2)
In ∆DEF and ∆PQR
∠D ≅ ∠Q
∠R ≅ ∠E
By AA test of similarity
∆DEF~ ∆PQR
DEPQ=EFQR=DFPR              Corresponding sides of similar triangles are proportional
DEPQEFRP
Hence, the correct option is (B).

(3)
In ∆ABC and ∆DEF 
∠B = ∠E,
∠F = ∠C
By AA test of similarity
∆ABC ~ ∆DEF
Since, there is not any congruency criteria like AA.
Thus, ∆ABC and ∆DEF are not congruent.
Hence, the correct option is (B).

(4)
Given: ∆ABC and ∆DEF are equilateral triangles
Constrcution: Draw a perependicular from vertex A and D on AC and DF in both triangles.
 
In ∆ABX and ∆DEY
∠B = ∠C = 60             (∆ABC and ∆DEF are equilateral triangles)
∠AXB = ∠DYB           (By construction)
By AA test of similarity
∆ABX ~ ∆DEY
ABDE=AXDY              Corresponding sides of similar triangles are proportional
DEPQEFRP

AABCADEF=1212×AB×AX12×DE×DY=12AB2DE2=12                    ABDE=AXDY DE2=32DE=42
Hence, the correct option is (D).

(5)
Given: seg XY || seg BC
By basic proportionality theorem
AXBX=AYYCBXAX+1=YCAY+1BX+AXAX=YC+AYAY
ABAX=ACAYABAC=AXAY
Hence, the correct option is (D).



Page No 27:

Question 2:

In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratios.



(1) AABDAADC

(2) AABDAABC

(3) AADCAABC

 

Answer:

Construction: Draw a perpendicular from vertex A to line BC.

(1) 
AABDAADC=12×AX×BD12×AX×DC=BDDC=713                     DC=BC-BD
(2) 
AABDAABC=12×AX×BD12×AX×BC=BDBC=720
(3) 
AADCAABC=12×AX×DC12×AX×BC=DCBC=1320                              DC=BC-BD

Page No 27:

Question 3:

Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle ?

Answer:

Area of  smaller triangleArea of bigger triangle=2312×Height of smaller triangle×Base of smaller triangle12×Height of bigger triangle×Base of bigger triangle=236Base of bigger triangle=23
Base of bigger triangle=32×6=9

Page No 27:

Question 4:

In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then A ABCA DCB=?

Answer:

Given:
∠ABC = ∠DCB = 90°
AB = 6
DC = 8
Now, A ABCA DCB=12×AB×BC12×DC×BC=68=34

Page No 27:

Question 5:

In the given figure, PM = 10 cm A(∆PQS) = 100 sq.cm A(∆QRS) = 110 sq.cm then Find NR.

Answer:

Given:
PM = 10 cm
A(∆PQS) = 100 sq.cm
A(∆QRS) = 110 sq.cm
Now, A PQSA QRS=10011012×PM×QS12×RN×QS=1011
10RN=1011RN=11 cm

Page No 27:

Question 6:

∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio AMNTAQRS.

Answer:

The areas of two similar triangles are proportional to the squares of their corresponding altitudes.
AMNTAQRS=592=2581



Page No 28:

Question 7:

In the given figure, A – D– C and B – E – C seg DE || side AB If AD = 5, DC = 3, BC = 6.4 then Find BE.

Answer:

Given:
AD = 5,
DC = 3,
BC = 6.4 
In △ABC,  DE || AB
CDDA=CEEB               By basic proportionality theorem35=6.4-xx3x=32-5x​​
8x=32x=4

Page No 28:

Question 8:

In the given figure, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70, CD = 80, PS = 280 then Find PQ, QR and RS.

Answer:

Given:
AB = 60,
BC = 70,
CD = 80,
PS = 280
Now, AD = AB + BC + CD
= 60 + 70 + 80
= 210
By intercept theorem, we have
PQAB=QRBC=RSCD=PSADPQ60=QR70=RS80=280210PQ60=QR70=RS80=43
PQ=43×60=80QR=43×70=2803RS=43×80=3203
 

Page No 28:

Question 9:

In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR.

Answer:

In △PMQ, ray MX is bisector of △PMQ.
PXXQ=MQMP                       .......... (I) theorem of angle bisector.
 In △PMR, ray MY is bisector of△PMR.
PYYR=MRMP                       .......... (II) theorem of angle bisector.
ButMPMQ=MPMR                      ......... M is the midpoint QR, hence MQ = MR.
PXXQ=PYYR 
∴XY || QR .......... converse of basic proportionality theorem.



Page No 29:

Question 10:

In the given fig, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find AXXY.

Answer:

In △ABY, ∠YBX = ∠XBA
AXXY=ABBY                     ...I  By angle bisector theorem
In △ACY, ∠YCX = ∠XCA
AXXY=ACCY                     ...II  By angle bisector theorem
From (I) and (II)
ACCY=ABBYACBC-BY=ABBY46-BY=5BY
4BY=30-5BY9BY=30BY=103
From (I), we have
AXXY=5103=32

Page No 29:

Question 11:

In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that APPD=PCBP

Answer:

Given: ▢ABCD is a parallelogram
To prove: APPD=PCBP
Proof: In △APD and △CPB
∠APD = ∠CPB         (Vertically opposite angles)
∠PAD = ∠PCB         (Alternate angles, AD || BC and BD is a transversal line)
By AA test of similarity
△APD ∼ △CPB 
APPC=PDPB                      Corresponding sides are proportionalAPPD=PCPB
Hence proved.

Page No 29:

Question 12:

In the given fig, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to Find the value of AC.

Answer:

Given:
XY || seg AC
2AX = 3BX
XY = 9
Consider, 2AX = 3BX
AXBX=32AX+BXBX=3+22                 .....by componendo
ABBX=52                   ....I       BCA~BYX         ...SAS test of similarityBABX=ACXY                 ...corresponding sides of similar triangles52=AC9    AC=17.5                   ...from I           

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Question 13:

In the given figure, the vertices of square DEFG are on the sides of ∆ABC. ∠A = 90°. Then prove that DE2 = BD × EC (Hint : Show that ∆GBD is similar to ∆CFE. Use GD = FE = DE.)

Answer:

Given: ▢DEFG is a square
To prove: DE2 = BD × EC
Proof: In △GBD and △AGF
∠GDB = ∠GAF =  90°         (Given)
∠AGF = ∠GBF                    (Corresponding, GF || BC and AB is a transversal line)
By AA test of similarity
△GBD ∼ △AGF                         ...(1)
In △CFEand △AGF
∠FEC = ∠GAF =  90°         (Given)
∠FCE = ∠AGF                   (Corresponding, GF || BC and AC is a transversal line)
By AA test of similarity
△CFE ∼ △AGF                          ...(2)
From (1) and (2), we get
△CFE ∼ △GBD     
CEGD=FEBD                      Corresponding sides are proportionalCEDE=DEBD                   GD=FE=DEDE2=BD×CE
Hence proved.



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