Question

If ABCD is a prallelogram, then prove the $area\left(△ABD\right)=area\left(△BCD\right)=area\left(△ABC\right)=area\left(△ACD\right)=\frac{1}{2}area\left(|{|}^{gm}ABCD\right)$

Answer 1

In || gm ABCD, BD and AC are joined

To Prove : area $\left(△ABD\right)=area(△BCD=area(△ABC)=area(△ACD)=\frac{1}{2}area(|{|}^{gm}ABCD)$

Proof: $\because$ Diagonals of a parallelogram bisect it into two triangles equal in area.

When BD is the diagonal, then,

area($△ABD)=area(△BCD)=\frac{1}{2}area(|{|}^{gm}ABCD)$ -----(i)

When AC is the diagonal, then,

area ($△ABC)=area(△ACD)=\frac{1}{2}area(|{|}^{gm}ABCD)$ ----(ii)

From (i) and (ii) ,

$area\left(△ABD\right)=area\left(△BCD\right)=area\left(△ABC\right)=area\left(△ACD\right)=\frac{1}{2}area\left(|{|}^{gm}ABCD\right)$