Question

In an equilateral triangle ABC , A is a point on side BC such that BD = 1/3 BC prove that 9(AD)^=7(AB)^

Answer 1

the answer is in this picture

Answer 2

hello here is ur answer hope it helps you
given: triangle ABC is an equilateral triangle. D is a point on BC such that BD=1/3BC
To prove:
${9ad}^{2} = {7ab}^{2}$
Construction: draw AM perpendicular to BC
Proof: AM perp. to BC
BM=CM
M is mid point of BC
In triangle AMD
${ad}^{2} = {am}^{2} + {md}^{2}$
${ad}^{2} = {am}^{2} + {(bm - bd)}^{2}$
${ad}^{2} = {am}^{2} + {bm}^{2} - 2bm \times bd + {bd}^{2}$
${ad}^{2} = {ab}^{2} - 2 \times \frac{1}{2} bc \times \frac{1}{3} bc + {( \frac{1}{3}bc) }^{2}$
${ad}^{2} = ab - { \frac{1}{3}bc }^{2} + { \frac{1}{9}bc }^{2}$
${ad}^{2} = {ab}^{2} - { \frac{1}{3}ab }^{2} + { \frac{1}{9}ab }^{2}$
${ad}^{2} = \frac{9 {ab}^{2} - 3 {ad}^{2} + {ab}^{2} }{9}$
$9 {ad}^{2} = 7 {ab}^{2}$
Hence Proved