Question

prove theorem 9.1 class 9​

Answer 1

$\huge\bold\green {HOLA!!}$

Parallelograms on the same base and between the same parallels

are equal in area.

Proof : Two parallelograms ABCD and EFCD, on

the same base DC and between the same parallels

AF and DC are given (see Fig.9.12).

We need to prove that ar (ABCD) = ar (EFCD).

In  ADE and  BCF,

 DAE =  CBF (Corresponding angles from AD || BC and transversal AF) (1)

 AED =  BFC (Corresponding angles from ED || FC and transversal AF) (2)

Therefore,  ADE =  BCF (Angle sum property of a triangle) (3)

Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)

So,  ADE   BCF [By ASA rule, using (1), (3), and (4)]

Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)

Now, ar (ABCD) = ar (ADE) + ar (EDCB)

= ar (BCF) + ar (EDCB) [From(5)]

= ar (EFCD)

So, parallelograms ABCD and EFCD are equal in area