Discussion Forum

 p₁,p₂,p₃  arte the altitudes  of a triangle ABC drawn from the vertices A,B and C respecively . If $\triangle$  is the area  of the triangle and 2s is the sum of its sides a,b and c , then $\frac{1}{p_{1}}+\frac{1}{p_{2}}-\frac{1}{p_{3}}$=

Answer:

Explanation:

Given

 BC=a,  AC-b

AB=c, AD=P₁

BE= p₂ , CF= p₃

 In $\triangle ABC$

       $\sin B=\frac {p_{1}}{c} \Rightarrow  \frac{1}{p_{1}}= \frac {1}{c \sin B}$

 Similarly  ,    $\frac{1}{p_{2}}=\frac{1}{a \sin C}$

and      $\frac{1}{p_{1}}=\frac{1}{b \sin A}$

$\therefore$     $\frac{1}{p_{1}}+\frac{1}{p_{2}}-\frac{1}{p_{3}}=\frac{1}{c\sin B}+\frac{1}{a\sin C}-\frac{1}{b\sin A}$

 =$\frac{a}{2 \triangle}+\frac{b}{2 \triangle}-\frac{c}{2 \triangle}=\frac{1}{2 \triangle}(a+b-c)=\frac {2(s-c)}{2 \triangle}=\frac{(s-c)}{\triangle}$

$\therefore$  $\frac{1}{p_{1}}+\frac{1}{p_{2}}-\frac{1}{p_{3}}$=  $\frac{s }{\triangle}$