IN equilateral triangle ABC if BC extended to D then prove that angle abc is greater than angle BAC
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Answer:
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
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