In parallelogram ABCD,two points P and Q are taken on diagonal BD such that DP = BQ show that 1.∆APD~= ∆ CQB,2.AP=CQ,3.∆AQB=∆CPD,4.AQ=CP,5.APCQ is a parallelogram
Answers
Answer:
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Step-by-step explanation:
Show that: (i) ΔAPD ≅ ΔCQB (ii) AP = CQ (iii) ΔABC ... P and Q are taken on diagonal BD such that DP = BQ (see Fig. ... iii ) ∆AQB congruent to ∆CPD iv ) AQ = CP
Answer:
Given :
In parallelogram ABCD , two points
P and Q are taken on diagonal BD
such that DP = BQ .
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To prove :
i ) ∆APD congruent to ∆CQB
ii ) AP = CQ
iii ) ∆AQB congruent to ∆CPD
iv ) AQ = CP
v ) APCQ is a parallelogram
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Construction : join AC to intersect
BD at O.
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Proof :
i ) In ∆APD and ∆CQB,
AD // BC
[ since , Opposite sides of parallelogram
ABCD and a transversal BD intersects
them ]
<ADB = <CBD
[ Since , Alternate interior angles ]
=> <ADP = <CBQ --------( 1 )
DP = BQ ------------( 2 ) [ given ]
AD = CB ------( 3 )
[ Opposite sides of parallelogram ABCD ]
In view of ( 1 ) , ( 2 ) and ( 3 )
∆APD congruent to ∆CQB
[ Since , SAS congruence criterion ]
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ii ) ∆APD congruent to ∆CQB
[ proved in ( i ) above ]
AP = CQ [ CPCT ]
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iii ) In ∆AQB and ∆CPD ,
AB // CD
[ Opposite sides of parallelogram
ABCD and a transversal BD intersects
them ]
<ABD = <CDB
[ Alternate interior angles ]
=> <ABQ = <CDP
QB = PD [ given ]
AB = CD
[ opposite sides of parallelogram
ABCD ]
∆AQB congruent to ∆CPD
[ SAS congruence Rule ]
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iv ) ∆AQB congruent to ∆CPD
proved in ( iii ) above
AQ = CP [ CPCT ]
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v ) The diagonals of a parallelogram
bisectors each other.
OB = OD
=> OB - BQ = OD - DP
[ Since BQ = DP given ]
OQ = OP -----( 1 )
OA = OC
[ Since , diagonals of a parallelogram
bisect each other ]
From , ( 1 ) and ( 2 ) ,
APCQ is a parallelogram.
I hope this helps you.
: )
Step-by-step explanation: