Question

In adjoining figure, $AD,BE$ & $CF$ are altitudes of $△ABC$. If $AD=CF=BE.$ Prove that $ABC$ is an equilateral triangle.

Answer 1

In$\Delta ADC$ and $\Delta BEC$

$\angle D=\angle E={90}^{\circ }$

$AD=BE$ (given)

$\angle C=\angle C$ (common)

$\therefore \Delta ADC\cong \Delta BEC$ by AAS property.

$AC=BC$ (by CPCT) ____ (1)

In $\Delta ADB$ and $\Delta CEB$

$\angle D=\angle F\left({90}^{\circ }\right),AD=CF$ (given), $\angle B=\angle B$

(common)

$\Delta ADB\cong \Delta CFB$ by AAS property

$AB=CB$ (by CPCT) _____ (2)

$AB=BC=CB$

Hence, $\Delta ABC$ is equilateral.