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Question (1):

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Answer:
AB = BC and AD = CE
AD = CE
AB=BC (data) AD=CE (data)
BD=BE (If equals are subtracted from equals, the remainders are also equal) In D AEB and D CDB, AB=BC (data) BE=BD (proved above)
(common angle to both the triangles)
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Question (2):
ABC and DEF are two triangles such that AB = DE, 

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Answer:
AB=BC and AD=CE

Comparing triangles ABC and DEF, AB=DE (data)
 BC = EF (data)
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Question (3):

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Answer:


Comparing D GCB and D DCE, BC=CE (data)



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Question (4):
ABC and ABD are two triangles such that AD=BC, BD=AC. 
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Answer:
In triangles ABC and ABD, AD=BC and BD=AC.

Comparing D ABC and D ABD , BC = AD (data) AC = BD (data) AB = AB (common side to both triangles)

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Question (5):

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Answer:
 AB = DE



BF = DG (corresponding sides) In D ABF and D DGE,
(data) AB = DE (data) BF = DG (from above)

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Question (6):

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Answer:
 GE = FA
AB = DE
(data)
BF = DG (corresponding sides of congruent triangles)

(supplements of equal angles are equal) In D ABF and D DGE, BF = DG (prove above) AF = GE (data)

 \ AB = DE (corresponding sides) |
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Question (7):

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Answer:
 GH = FH
BG = DF

BC = CD BF = GD (corresponding sides of congruent triangles)
(corresponding angles of congruent triangles) Compare D GBH and D DHF, GH = FH (data) BF+FH = GD+GH (BF=GD and FH=GH) i.e., BH = DH
(proved above)

BG = DF (congruent parts of congruent triangles) |
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Question (8):

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Answer:
QR = SR


Compare D TQR and D TSR,
 QR = SR (data) TR = TR (common side)


 QT = TS (CPCT) Now compare D PTQ and D PTS,
 TQ = TS (prove above) PT = PT (common side)

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Question (9):

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Answer:


CF=BE
Comparing D FAC and D BED,
(data)
(data) AC = BD (AB=CD given. Add BC to both sides. We get AB+BC = CD+BC i.e., AC=BD)

CF=BE ( CPCT) |
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Question (10):
In the following examples, the hypothesis is marked on the diagram. Write the postulate of congruence you would use to support your argument that the triangles are congruent. If more than one hypothesis is applicable to any example, state both. |
Answer:
i)
Postulate : SAS ii)
Postulate : SSS, SAS iii)
Postulate : ASA iv)
Postulate : SSS v)
Postulate : SAS |
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Question (11):
In the figure BA=BC, prove that AD = CE.
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Answer:
BA = BC, CE ^ AB, AD ^ BC
AD = CE
In triangles AEC and ADC,
( Given) AC=AC ( common side)
( AB=BC) (given)


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Question (12):
In the adjoining figure, if BAC and DEF are right angles triangles, prove that AB = EF.
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Answer:
In triangles ABC and DEF, AC=DE

AB=EF
In triangles DEF and ABC,
 AC=DE ( given)
( given)
(ASA Postulate)
Hence, AB = EF (c.p.c.t) |
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Question (13):
Select congruent triangles in the following set of triangles and state the congruency condition in each case.

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Answer:
Case I:
 In D III right angle is not formed between 3 and 4.
They are not congruent triangles Case II:
 In D I side 3 is not the included side.
DI is not congruent to DII and DIII. |
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Question (14):
ABCD is a quadrilateral in which DF and BE are perpendicular drawn on the diagonal AC. If AE=FC and BC=AD, prove that BE=FD.
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Answer:
ABCD is a quadrilateral, diagonal AC is drawn BE ^ AC and DF ^ AC. AE=FC and BC=AD.
BE=FD
Compare triangles BEC and AFD, Hyp. BC= hyp. AD (given)
 AE=FC ( given)
AC-AE=AC-FC i.e., EC=AF

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Question (15):
In the adjoining figure, O is the mid-point of AB and CD. Prove that AC=BD and AC||BD.
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Answer:
 AC and BD are joined.
AC=BD
In triangles AOC and BOD,




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Question (16):
In the adjoining figure, AB and CD intersect at O such that AO=OD and OB=OC. Prove that AC=BD.
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Answer:
AB and CD intersect at O such that AO=OD and OB=OC. AC and BD are joined.

In triangles AOC and BOD,




AC=BD |
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Question (17):
In the adjoining figures, two sides AB and BC and the median AD of D ABC are equal respectively to the two sides PQ and

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Answer:
In triangles ABC and PQR,




In triangles ABD and PQM AB=PQ ( given) AD=PM ( given) BD=QM (BC=QR and AD and PM are medians of BC and QR)







  (When equals are subtracted from equals the retaining parts are equal) Now compare triangles ADC and PMR,
(given) DC = MR (BC=QR) (given)
( Proved above)
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Question (18):
In the adjoining figure, AB=AC, BE and CF are respectively,

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Answer:


In triangles EBC and FCB, BC=BC ( common side)
(AB=AC given) ( base angles are equal)


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Question (19):
In the adjoining figures, AB=AC and DB=DC. Prove that .
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Answer:
With reference to both fig(a) and (b)


Join AD.
Comparing triangles ABD and ADC,




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Question (20):
D ABC is an isosceles triangle with AB=AC side BA is produced

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Answer:
In D ABC, AB=AC, DA is produced such that AB=AD.

In D ABC, AB = AC ( given)
( base angles of an isosceles triangle are equal)
 
 


 In D ADC,
AC=AD (given)
  ( base angles of an isosceles triangle)
 (angle sum property)




 

 Adding (i) and (ii),

 = 180 - 90 = 90o

 
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