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Given ΔABC is equilateral and AD⊥BC. That is AD bisects BC at 'D'.
It is given that the length of the side BC is 'x' units.⇒AB=BC=AC=x units.
Here, ΔABD is right angled triangle.
So,by Pythagoras theorem,
AB²=AD²+BD²
⇒x²=AD²+(x/2)² [∵BC is bisected at D]
⇒AD²=x²-x²/4
⇒AD²=3x²/4
⇒AD=[(√3)x]/2.
It is given that the length of the side BC is 'x' units.⇒AB=BC=AC=x units.
Here, ΔABD is right angled triangle.
So,by Pythagoras theorem,
AB²=AD²+BD²
⇒x²=AD²+(x/2)² [∵BC is bisected at D]
⇒AD²=x²-x²/4
⇒AD²=3x²/4
⇒AD=[(√3)x]/2.
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