Theorem 9.1 : Parallelogram on the same base and between the same parallels are equal in area Class 9

Given : ABCD and EFCD are two parallelograms that have the same base CD and lie between the same parallels AF and CD.

To prove : ar(ABCD) = ar(EFCD)

Proof : opposite sides of a parallelogram are parallel

⇒ AD || BC

With transversal AB

⇒ ∠DAE = ∠CBF (Corresponding angle)

⇒ AD = BC (opposite sides  are equal in parallelogram)

⇒ ED || FC

With transversal EF

⇒ △AED ⩭ △BPC (by ASA congruency)

The area of the two congruent figures is the same. adding ar (EBCD) on both sides

⇒ (AED) + ar(EBCD) = ar(BFC) + ar(EBCD)

⇒ ar(ABCD) = ar(EFCD)

Hence proved.

Theorem 10.3 : The perpendicular from the center of a circle to a chord bisects the chord Class 9

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