Question 15
In an equilateral triangle ABC, D is a point on side BC such that BD=13BC. Prove that 9AD2=7AB2.
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Solution
Let the side of an equilateral triangle bea, and AE be the altitude of ΔABC. ∴BE=EC=BC2=a2
And, AE=a√32
given that, BD=13BC ∴BD=a3 DE=BE−BD=a2−a3=a6
Applying Pythagoras theorem in ΔADE, we get AD2=AE2+DE2 ⇒AD2= (a√32)2+(a6)2 ⇒AD2=(3a24)+(a236) ⇒AD2=28a236 ⇒AD2=79AB2 [Since, the side of an equilateral triangle is a.i.e;AB=BC=AC=a] ⇒9AD2=7AB2