यदि `p_(1), p_(2),p_(3)` किसी त्रिभुज ABC के शीर्ष A, B, C से क्रमशः विपरीत भुजा | पर लम्ब पर लम्बाई हो तथा त्रिभुज ABC का क्षेत्रफल `Delta` हो , तो साबित करे कि
`(1)/(P_(1)) + (1)/(P_(2)) -(1)/(P_(3)) = (2ab cos^(2)""(C)/(2))/(triangle(a+b+c))`

Explanation

`because Delta = (1)/(2). a.p_(1)`
`therefore (1)/(p_(1)) = (a)/(2Delta)`
उसी तरह, ` (1)/(p_(2)) = (b)/(2Delta)`
तथा ` (1)/(p_(3)) = (c)/(2Delta)`
अब दायाँ पक्ष ` = (2)/(2Delta) + (b)/(2Delta) - (c)/(2Delta) = (a + b +c)/(2Delta)`
` = (a + b -c)/(2Delta) * (a + b + c)/(a + b +c ) = ((a + b)^(2) - c^(2))/(2Delta(a + b +c))`
` = (2ab + a^(2) + b^(2) -c^(2))/(2Delta(a + b+c)) = (2ab +2ab cosC)/(2Delta(a + b+c))[because cosC = (a^(2) + b^(2) -c^(2))/(2ab)]`
` = (2ab(1 + cosC))/(2Delta(a + b+c)) = (ab 2cos^(2)C//2)/(Delta(a+b+c)) = (2ab cos^(2)""(C)/(2))/(Delta(a+b+c)) =` दायाँ पक्ष