Question

Prove that the area of triangle standing on the same base or equal bases and between same parallels are equal in area.

Answer 1

Draw a line through A parallel to BC meets DC at E, i.e., AE$\parallel$BC and a line through B parallel to AD meets DC at F, i.e, BF$\parallel$AD.

In ABCD,

DF$\parallel$AB (Given, as AB$\parallel$CD) & BF$\parallel$AD (By contruction)

$\therefore$ Both pair of opposite sides are parallel. Hence, ABFD is parallelogram. Similarly in ABCE, EC$\parallel$AB (Given, AB$\parallel$CD) & AE$\parallel$BC (By construction). Thus, both pair of opposite sides are parallel, hence ABCE is a parallelogram.

Therefore, ABFD and ABCE are two parallelogram w it sam base AB and between two $\parallel$als AB & EF.

$\therefore$ ar (ABFD) = ar (ABCE) $⟶\left(1\right)$

In $\parallel$gm ABFD, diagonals divides it into two congruent $△B$

$△ABD\cong △FBD$, Similarly in $\parallel$gm ABCE, $△ABC\cong △AEC$ And area of congruent triangles are equal, thux ar $△$ABD = ar $△$FBD & ar $△$ABC = ar $△$AEC

$\Rightarrow$ ar (ABFD) = 2ar ($△$ABD) $⟶\left(2\right)$

$\Rightarrow$ ar (ABCE) = 2ar ($△$ABC) $⟶\left(3\right)$

From equation (1), ar (ABFD) = ar (ABCF)

$\Rightarrow$ 2ar ($△$ABD) = 2(ar $△$ABC)

$\Rightarrow$ ar $△$ABD = ar $△$ABC

Hence Proved.