Question

Prove that all triangles with the same base and third vertex on a line parallel to the base have same area.?​

Answer 1

Answer:

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

In ABCD,  

DF∥AB (Given, as AB∥CD) & BF∥AD (By contruction)

∴  Both pair of opposite sides are parallel. Hence, ABFD is parallelogram. Similarly in ABCE, EC∥AB (Given, AB∥CD) & AE∥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 ∥als AB & EF.

∴  ar (ABFD) = ar (ABCE)   ⟶(1)

In ∥gm ABFD, diagonals divides it into two congruent △B

△ABD≅△FBD, Similarly in  ∥gm ABCE, △ABC≅△AEC And area of congruent triangles are equal, thux ar △ABD = ar △FBD & ar △ABC = ar △AEC

⇒      ar (ABFD) = 2ar (△ABD)   ⟶(2)

⇒      ar (ABCE) = 2ar (△ABC)    ⟶(3)

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

⇒      2ar (△ABD) = 2(ar △ABC)

⇒      ar △ABD = ar △ABC

Step-by-step explanation: