Question

in triangle ABC, d and e are mid points of sides ab and ac respectively.a line through c parallel to bd meets de produced in f.show that ar(abc):ar(bcfd)=1:1

Answer 1

Answer:

Step-by-step explanation:

In ΔABC, D and E are the midpoints of AB and AC respectively.

Therefore, DE II BC (By Converse of mid-point theorem)

Also, DE = 1/2BC

In ΔADE and ΔABC

∠ADE = ∠B   (Corresponding angles)

∠DAE = ∠BAC   (common)

ΔADE–ΔABC (By AA Similarity)

We know that the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

ar(ΔADE)/ar(ΔABC) = AD²/AB²

ar(ΔADE)/ar(ΔABC) = AD² / 2AD²

[AD = 2AB as D is the mid point]

ar(ΔADE)/ar(ΔABC) = 1² /2²

ar(ΔADE)/ar(ΔABC) = 1/4

Hence, the ratio of the areas ΔADE and ΔABC is ar(ΔADE) : ar(ΔABC) = 1 : 4.