Question

P is the midpoint of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R. Prove that
AR = 2BC and BR = 2BQ

(With diagram)

Answer 1

Hello Mate!

Given : P is the midpoint of side AB of a || gm ABCD. A line through B parallel to PD meets DC at Q and AD produced at R.

To prove : AR = 2BC and BR = 2BQ

Proof : Since PD || BQ or BR where P is mid point at AB therefore by theorum we get that,

D is also mid mid point on side AR of ∆ABR.

Hence, AD = DR

AD + DR = AR

AD + AD = AR

2AD = AR.

Since AD = BC ( opp. sides of ||gm are equal )

Therefore, 2AD = AR.

Now, in ∆ABR only, DQ || AB ( ||gm property ).

Where D is mid point on AR so Q is mid point at BR.

So, BQ = QR

BQ + QR = BR

2BQ = BR.

Have great future ahead!

Answer 2

Heya !

Here's your answer !!

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GIVEN :

ABCD is a ||gm

P is mid point of AB

PD || BR

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R.T.P. :

( i ) AR = 2BC

( ii ) BR = 2 BQ

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PROOF :

( i ) In triangle ARB ,

P is mid point of AB and PD || BR

Therefore,

D is the mid pint of AR

{ Converse of Mid point Theoram }

=> AR = 2AD

=> AR = 2BC { AD = BC }

[ HENCE PROVED ]


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( ii ) DC || AB

=> DQ || AB

In triangle RAB

D is mid point of AR and DQ || AB

Therefore,

Q is mid point of RB

{ Converse of Mid point Theoram }

=> BR = 2BQ

[ HENCE PROVED ]


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THANKS !!