Example 2 of chapter 9 areas …

{FOR DIAGRAM SEE IN EXAMPLE 2 } 1• At first we draw BQ parallel AP so that ABQP is a parallelogram in which AP||BQ & AB||PQ 2• now both parallelograms ABQP & ABCD are equal in area , ar(ABQP)=ar(ABCD) _____________(1) 3• In parallelogram ABQP ; is a diagonal so ∆PAB≈∆BQP ; { The diagonal of a parallelogram divides it into two congruent triangles } So, ar(PAB) = ar(BQP) ; _________(2) Because the triangles which are congurent are equal in area's. 4• so From Eq.2 ar(PAB) =1/2 ar(ABQP) _________(3) 5• So From Eq.1&3 we get -: ar(PAB) =1/2 ar(ABCD) (Hence Proof)