Let $C$ be incircle of $△ A B C$ . If tangents are drawn parallel to the sides the $△$ such that $t_{1}, t_{2}, t_{3}$ be the lengths of tangents which lie inside the given triangle parallel to sides $a, b, c$ respectively, then $\frac{t_{1}}{a} + \frac{t_{2}}{b} + \frac{t_{3}}{c}$ is equal to ?

= 0

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= 1

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= 2

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= 3

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Solution

The correct option is A $= 0$
In the given figure,
$C$ is the incircle of $△ A B C$ .
Tangent $t_{1}$ is parallel to side a of triangle, tangent $t_{2}$ is parallel to side b of triangle and tangent $t_{3}$ is parallel to side c of triangle.

Refer to the figure. Consider $△ A P_{1} P_{2}$ and $△ A B C$

By $A A A$ test, $△ A P_{1} P_{2} \sim △ A B C$
Thus, by property of similar triangles,

$\frac{t_{1}}{a} = \frac{h - 2 r}{h}$

$∴ \frac{t_{1}}{a} = 1 - \frac{2 r}{h}$

But, radius of incircle is, $r = \frac{△}{s}$

$∴ \frac{t_{1}}{a} = 1 - \frac{2}{h} \times \frac{△}{s}$
where, $△$ = area of $△ A B C$
$s$ = semi perimeter of triangle

$∴ \frac{t_{1}}{a} = 1 - \frac{2}{h} \times \frac{\frac{1}{2} \times a \times h}{s}$

$∴ \frac{t_{1}}{a} = 1 - \frac{a}{s}$ (1)

Similarly, $\frac{t_{2}}{b} = 1 - \frac{b}{s}$ (2)

$\frac{t_{3}}{c} = 1 - \frac{c}{s}$ (3)

Adding equations (1), (2) and (3), we get,

$\frac{t_{1}}{a} + \frac{t_{2}}{b} + \frac{t_{3}}{c} = 1 - \frac{a}{s} + 1 - \frac{b}{s} + 1 - \frac{c}{s}$

$∴ \frac{t_{1}}{a} + \frac{t_{2}}{b} + \frac{t_{3}}{c} = 3 - \frac{a + b + c}{s}$

Now, $s = \frac{a + b + c}{2}$

$∴ \frac{t_{1}}{a} + \frac{t_{2}}{b} + \frac{t_{3}}{c} = 3 - \frac{a + b + c}{(\frac{a + b + c}{2})}$

$∴ \frac{t_{1}}{a} + \frac{t_{2}}{b} + \frac{t_{3}}{c} = 3 - 2$

$∴ \frac{t_{1}}{a} + \frac{t_{2}}{b} + \frac{t_{3}}{c} = 1$

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Let C be incircle of △ABC. If tangents are drawn parallel to the sides the △ such that t1,t2,t3 be the lengths of tangents which lie inside the given triangle parallel to sides a,b,c respectively, then t1a+t2b+t3c is equal to ?