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Let ABC be an equilateral triangle and P...

Let ABC be an equilateral triangle and P be a point within the triangle. Perpendicular PD, PE and PF are drawn to three sides of the triangle, then the value of `(AB+BC+CA)/(PD+PE+PF)` is

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BD is the disector of angle ABC. From a point P in BD, perpendiculars PE and PF are drawn to AB and BC respectively, prove that : (i) Triangle BEP is conguent to triangle BFP (ii) PE=PF.

BD is the disector of angle ABC. From a point P in BD, perpendiculars PE and PF are drawn to AB and BC respectively, prove that : (i) Triangle BEP is conguent to triangle BFP (ii) PE=PF.