If ABCD is a parallelogram, then prove that ar(Δ ABD) = ar(Δ BCD) =

1.	If ABCD is a parallelogram, then prove that ar(Δ ABD) = ar(Δ BCD) = ar(Δ ABC) = ar(Δ ACD) = ar(\|\|gm ABCD)
Answer» We know that, Diagonal of parallelogram divides it into two quadrilaterals. Since, AC is the diagonal Then, Area (ΔABC) = Area (ΔACD) = \(\frac{1}{2}\) Area of parallelogram ABCD ...(i) Since, BD is the diagonal Then, Area (ΔABD) = Area (ΔBCD) = \(\frac{1}{2}\)Area of parallelogram ABCD ...(ii) Compare (i) and (ii), we get Therefore, Area (ΔABC) = Area (ΔACD) = Area (ΔABD) = Area (ΔBCD) = \(\frac{1}{2}\)Area of parallelogram ABCD

If ABCD is a parallelogram, then prove that ar(Δ ABD) = ar(Δ BCD) = ar(Δ ABC) = ar(Δ ACD) = ar(||gm ABCD)

Answer»

We know that,

Diagonal of parallelogram divides it into two quadrilaterals.

Since,

AC is the diagonal

Then,

Area (ΔABC) = Area (ΔACD)

= \(\frac{1}{2}\) Area of parallelogram ABCD ...(i)

Since,

BD is the diagonal

Then,

Area (ΔABD) = Area (ΔBCD)

= \(\frac{1}{2}\)Area of parallelogram ABCD ...(ii)

Compare (i) and (ii), we get

Therefore,

Area (ΔABC) = Area (ΔACD)

= Area (ΔABD) = Area (ΔBCD)

= \(\frac{1}{2}\)Area of parallelogram ABCD