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