Question

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Answer 1

Answer:

In △BAD, P and S are mid-points of AB and BD

⇒ PS∥AD and PS=

2

1

AD --- ( 1 ) [ By BPT ]

Similarly in △CAD,

⇒ OR∥AD and QR=

2

1

AD --- ( 2 )

From ( 1 ) and ( 2 ), we get

⇒ PS∥QR and PS=QR=

2

1

AD ------ ( 3 )

In △BDC, we get

⇒ SR∥BC and SR=

2

1

BC ----- ( 4)

And in △ABC

PQ∥BC and PQ=

2

1

BC ------ ( 5 )

⇒ PQ∥SR, PQ=SR=

2

1

BC ---- ( 6 ) [ From ( 4 ) and ( 5 ) ]

∴ □PQRS is a parallelogram.

Now, AD=BC

∴

2

1

AD=

2

1

BC

∴ PS=QR=PQ=SR [ From ( 3 ) and ( 6 ) ]

∴ □PQRS is a rhombus.

HOPE IT IS HELPFUL TO U AND ALL

Answer 2

Answer:

1. since in both ∆ ab=ba (common arm), and ad=bc(given) and angle dab =angle cba(given)

therefore traingle abd and bac are congruent by sas congruency.

2. bd =ac (by cpct )

3. by cpct