Question

class 8 in the adjoining figure ,abcd is a parallelogram . perpendicular DN and BP are drawn on diagonal AC. Prove that​

Answer 1

$\huge\underline\mathfrak{Question-}$

In the adjoining figure ,abcd is a parallelogram . perpendicular DN and BP are drawn on diagonal AC.

Prove that ;-

(i) ∆DCN ≈ ∆BAP
(ii) AN = CP

$\huge\underline\mathfrak{Answer-}$

Here , since ;
ABCD is a parallelogram , than AB || CD & AD || DC.

Since , AB || DC & AC is a transversal.
/_ PAB = /_ NCD ( Alternate Interior Angles )

In ∆DCN & ∆BAP ;-
∆DCN = ∆BAP ( 90° each )
/_ NCD = /_ PAB ( Proved Above!)
DC = AB ( Opposite Sides of ||gm are equal )

=> ∆DCN ≈ ∆BAP ( AAS )
=> DN = BP ( CPCT )

In ∆DNA & ∆BPC ;-
AD = BC ( Opposite Sides of a ||gm are equal )
DN = BP ( Proved Above! )

=> ∆DNA ≈ ∆BPC ( RHS )
=> AN = CP ( CPCT )

Hence , Proved !

Answer 2

Heyya mate.....

Here , since ;

ABCD is a parallelogram , than AB || CD & AD || DC.

Since , AB || DC & AC is a transversal.

angle PAB = angle NCD ( Alternate Interior Angles )

In ∆DCN & ∆BAP ;-

∆DCN = ∆BAP ( 90° each )

angle NCD = angle PAB ( Proved Above!)

DC = AB ( Opposite Sides of ||gm are equal )

=> ∆DCN ≈ ∆BAP ( AAS )

=> DN = BP ( CPCT )

In ∆DNA & ∆BPC ;-

AD = BC ( Opposite Sides of a ||gm are equal )

DN = BP ( Proved Above! )

=> ∆DNA ≈ ∆BPC ( RHS )

=> AN = CP ( CPCT )

Hence , Proved !