Question

In triangle, AD perpendicular to BC, SC=1/3BC. triangle is an equilateral triangle prove that 9AD square=7AC square​

Answer 1

ABC be an equilateral triangle and D be the point on BC such that

BD=31BC (Given)

Draw AE⊥BC, Join AD.

BE=EC (Altitude drawn from any vertex of an equilateral triangle bisects the opposite side)

So, BE=EC=2BC

In ΔABC

AB2=AE2+EB2....(i)

AD2=AE2+ED2....(ii)

From (i) and (ii)

AB2=AD2−ED2+EB2

⇒AB2=AD2−36BC2+4BC2

(∵BD+DE=2BC⇒3BC+DE=2BC⇒DE=6BC)

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Answer 2

Answer:

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