SATHEE: Properties of Triangles 2 Question 15

Properties of Triangles 2 Question 15

15. If $p_{1}, p_{2}, p_{3}$ are the altitudes of a triangle from the vertices $A, B, C$ and $Δ$ is the area of the triangle, then prove that

$\frac{1}{p_{1}} + \frac{1}{p_{2}} - \frac{1}{p_{3}} = \frac{2 a b}{(a + b + c) Δ} \cos^{2} \frac{C}{2}$

(1978, 3M)

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Solution:

Since, $Δ = \frac{1}{2} a p_{1} \Rightarrow \frac{1}{p_{1}} = \frac{a}{2 Δ}$

Similarly, $\frac{1}{p_{2}} = \frac{b}{2 Δ}, \frac{1}{p_{3}} = \frac{c}{2 Δ}$

$\begin{aligned} ∴ \frac{1}{p_{1}} + \frac{1}{p_{2}} - \frac{1}{p_{3}} & = \frac{1}{2 Δ} (a + b - c) \\ = \frac{2 (s - c)}{2 Δ} = \frac{s - c}{Δ} = \frac{s (s - c)}{a b} \cdot \frac{a b}{s Δ} \\ = \frac{a b}{\frac{a + b + c}{2} Δ} \cdot \cos^{2} \frac{C}{2} \\ = \frac{2 a b}{(a + b + c) Δ} \cos^{2} \frac{C}{2} \end{aligned}$

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Properties of Triangles 2 Question 15

15. If p1,p2,p3 are the altitudes of a triangle from the vertices A,B,C and Δ is the area of the triangle, then prove that

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15. If $p_{1}, p_{2}, p_{3}$ are the altitudes of a triangle from the vertices $A, B, C$ and $Δ$ is the area of the triangle, then prove that