In an equilateral $\triangle ABC$, $AD \perp BC$, prove that $AD^2=3BD^2$.
Given:
In an equilateral $\triangle ABC$, $AD \perp BC$.
To do:
We have to prove that $AD^2=3BD^2$.
Solution:
In $\triangle ADB$ and $\triangle ACD$,
$\angle ADB=\angle ADC=90^o$
$AB=AC$
$AD=AD$ (Common side)
Therefore,
$\triangle ADB \cong\ \triangle ACD$ (By RHS congruence)
This implies,
$BD=DC=\frac{BC}{2}$ (CPCT)
In $\triangle ADB$,
$AB^2=AD^2+BD^2$ (By using Pythagoras theorem)
$BC^2=AD^2+BD^2$ (Since $AB=AC$)
$(2BD)^2=AD^2+BD^2$
$4BD^2=AD^2+BD^2$
$(4-1)BD^2=AD^2$
$AD^2=3BD^2$
Hence proved.
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