Question

In an equilateral triangle ABC the side BC is trisected at D prove that 7BC2 = 9AD2
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Answer 1

Step-by-step explanation:

Let side =a

Draw AE⊥BC So, as triangle ABC is equilateral.

BE=2a and as BD=3a

DE=BE−BD=6a

In △ADE by Pythagoras theorem

AD2=AE2+DE2

=(23a)2+36a2=97a2=97AB2

⇒9AD2=7AB2

Hence provef

Answer 2

Answer:

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