Question

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P such that ar(∆APD)=ar(∆BPC). Prove that ABCD is a trapezium.​

Answer 1

Step-by-step explanation:

since the ar (triangleAPD)=ar(TRIANGLE BPC)

• AB PARALLEL to CB

so

ABCD IS TRAPEZIUM.

(BECAUSE IN TRAPEZIUM TWO SIDES ARE PARALLEL)

Answer 2

Step-by-step explanation:

let the ABCD be a quadrilateral and AC and BD are its diagonal iintersectingat p.

given:triangle apd=triangle bpd

angle dap=angle pcb(cpct)

but these are alternate interior angle

so Ab//Bc

so ABCD is a trapezium