Question

# In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Solution

## Given:$$\triangle ABC$$ is an Equilateral Triangle.$$AB=BC=AC$$$$CD$$ Perpendicular to $$AB$$To Prove:$$3AC^2=4CD^2$$Proof:In $$\triangle ADC$$By Pythagoras Theorem,$$AC^2=AD^2+DC^2\,.....(1)$$In $$\triangle BDC$$By Pythagoras Theorem,$$BC^2=DC^2+BD^2\,.....(2)$$Adding 1) and 2) We get,$$2AC^2=2DC^2+AD^2+BD^2$$ $$(Since\,\, AC=BC)$$Since, $$AD=BD=\dfrac{1}{2}AC$$ $$(Since\,\,AC=AB)$$$$\therefore$$ $$2AC^2=2DC^2+\dfrac{1}{2}AC^2$$$$\therefore$$ $$2AC^2-\dfrac{1}{2}AC^2=2DC^2$$$$\therefore$$ $$3AC^2=4CD^2$$MathematicsRS AgarwalStandard X

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