The base BC of triangle ABC is divided at D so that BD = 1/2 DC. Prove that the area of Δ ABD = 1/3 of the area of Δ ABC.

Question

KnowledgeBoat · Accepted Answer

In △ ABC, ⇒ BD = DC ⇒ We know that, Ratio of the area of triangles with same vertex and bases along the same line is equal to the ratio of their respective bases. From figure, ⇒ Area of Δ ABC = Area of Δ ABD + Area of Δ ADC ⇒ Area of Δ ABC = Area of Δ ABD + 2 Area of Δ ABD ⇒ Area of Δ ABC = 3 Area of Δ ABD ⇒ Area of Δ ABD = Area of Δ ABC. Hence, proved that area of Δ ABD = area of Δ ABC.

Mathematics

The base BC of triangle ABC is divided at D so that BD = 1/2 DC. Prove that the area of Δ ABD = 1/3 of the area of Δ ABC.

Theorems on Area

Related Questions

In a parallelogram ABCD, point P lies in DC such that DP : PC = 3 : 2. If area of Δ DPB = 30 sq.cm, find the area of the parallelogram ABCD.

The median of a triangle divides it into two:
triangles of equal area
congruent triangles
right triangles
isosceles triangles

Mathematics

The base BC of triangle ABC is divided at D so that BD = 1/2 DC. Prove that the area of Δ ABD = 1/3 of the area of Δ ABC.

Theorems on Area

Related Questions

In a parallelogram ABCD, point P lies in DC such that DP : PC = 3 : 2. If area of Δ DPB = 30 sq.cm, find the area of the parallelogram ABCD.

The median of a triangle divides it into two:triangles of equal areacongruent trianglesright trianglesisosceles triangles

The median of a triangle divides it into two:
triangles of equal area
congruent triangles
right triangles
isosceles triangles