Question

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that: (i) ΔAPD ≅ ΔCQB (ii) AP = CQ (iii) ΔABC ≅ ΔCPD (iv) AQ=CP (v)APCQ is a parallelogram

Answer 1

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.

: )

Answer 2

Given ABCD is a parallelogram with points P and Q on diagonal BD such that DP = BQ.

(i)

In ΔAPD and ΔCQB, we have

⇒ AD = BC

⇒ DP = BQ

⇒ ∠ADP = ∠CBQ

⇒ ΔAPD ≅ ΔCQB {By SAS criterion of congruence}

(ii)

From (i)

AP = CQ

(iii)

In ΔAQB and ΔCPD.

⇒ ∠ABD = ∠CDP

⇒ ∠ABQ = ∠CDP

⇒ ΔABQ ≅ CPD.

(iv)

from (iii)

⇒ AQ = CP {SAS rule}

(v)

Given, BQ = DP

∴ OB = OD

⇒ OB - BQ = OD - DP

⇒ OQ = OP  ---- (i)

⇒ OA = OC  ---- (ii)

From (i) & (ii),

⇒ APCQ ia a parallelogram.

Hope it helps!

⇒ BQ = DP

⇒