SATHEE: Properties of Triangles 1 Question 23

Properties of Triangles 1 Question 23

23. $A B C$ is a triangle. $D$ is the middle point of $B C$ . If $A D$ is perpendicular to $A C$ , then prove that

$\cos A \cos C = \frac{2 (c^{2} - a^{2})}{3 a c}$

$(1980, 3 M)$

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Solution:

In $△ A D C$ , we have

Applying cosine formula in $△ A B C$ , we have

$\begin{aligned} \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \\ \cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} \end{aligned}$

From Eqs. (i) and (ii),

$\begin{aligned} \frac{a^{2} + b^{2} - c^{2}}{2 a b} & = \frac{2 b}{a} \\ \Rightarrow & a^{2} + b^{2} - c^{2} & = 4 b^{2} \\ \Rightarrow & a^{2} - c^{2} & = 3 b^{2} \end{aligned}$

Now, $\cos A \cos C = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \cdot \frac{2 b}{a}$

$\begin{aligned} = \frac{b^{2} + c^{2} - a^{2}}{a c} = \frac{3 b^{2} + 3 (c^{2} - a^{2})}{3 a c} \\ = \frac{(a^{2} - c^{2}) + 3 (c^{2} - a^{2})}{3 a c} = \frac{2 (c^{2} - a^{2})}{3 a c} \end{aligned}$

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Properties of Triangles 1 Question 23

23. ABC is a triangle. D is the middle point of BC. If AD is perpendicular to AC, then prove that

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23. $A B C$ is a triangle. $D$ is the middle point of $B C$ . If $A D$ is perpendicular to $A C$ , then prove that