Question

In Δ ABC, D and E are mid points of sides AB and AC respectively. Show that
ar(Δ ADE) = 1/4 ar(Δ ABC)

Answer 1

Given: D and E are mid points of AB and AC

By mid point theorem, DE∥BC

In △ADE and △ABC

∠DAE=∠BAC (Common)

∠ADE=∠ABC (Corresponding angles)

∠AED=∠ACB (Corresponding angles)

Thus, △ABC∼△ADE (AAA rule)

Hence,

A(△ABC)

A(△ADE) = AB 2AD 2 (Similar triangle property)A(△ABC)

A(△ADE) = (2AD) 2AD 2

A(△ABC)

A(△ADE) =1:4

Answer 2

Answer:

Given: D and E are mid points of AB and AC

By mid point theorem, DE∥BC

In △ADE and △ABC

∠DAE=∠BAC (Common)

∠ADE=∠ABC (Corresponding angles)

∠AED=∠ACB (Corresponding angles)

Thus, △ABC∼△ADE (AAA rule)

Hence,

A(△ABC)

A(△ADE)

=

AB

2

AD

2

(Similar triangle property)

A(△ABC)

A(△ADE)

=

(2AD)

2

AD

2

A(△ABC)

A(△ADE)

=1:999999999