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In an equilateral triangle ABC,AD is drawn perpendicular to BC meeting BC in D. Prove that AD sure = 3BD squre

Posted by Shyam Yadav 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

According to the question, {tex}\triangle ABC {/tex} is an equilateral triangle.
In {tex}\triangle{/tex}ABD, using Pythagoras theorem,

{tex}\Rightarrow{/tex} AB² = AD² + BD²
{tex}\Rightarrow{/tex} BC² = AD² + BD², (as AB = BC = CA)
{tex}\Rightarrow{/tex} (2 BD)² = AD² + BD², (perpendicular is the median in an equilateral triangle)
{tex}\Rightarrow{/tex} 4BD² - BD² = AD²
{tex}\therefore{/tex} 3BD² = AD²

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