In an equilateral triangle ABC the side BC is trisected at D prove that 7BC^2 = 9AD^2
Answers
Answered by
1
Answer:
Step-by-step explanation:
Let side =a
Draw AE⊥BC So, as triangle ABC is equilateral.
BE=
2
a
and as BD=
3
a
DE=BE−BD=
6
a
In △ADE by Pythagoras theorem
AD
2
=AE
2
+DE
2
=(
2
3
a)
2
+
36
a
2
=
9
7a
2
=
9
7AB
2
⇒9AD
2
=7AB
2
Similar questions