$A D$ bisects the $∠ A$ of a $Δ A B C$ and meets $B C$ at $D$ . If $B C = K$ metre, $C A = l$ metre and $A B = m$ metre, then find the length of $D C$ .

\frac{K l}{m + l}

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\frac{K (m + l)}{l}

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\frac{K m}{m + l}

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\frac{K (m + l)}{m}

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Solution

The correct option is A $\frac{K l}{m + l}$
By angle bisector theorem, we can write
$\frac{B D}{D C} = \frac{A B}{A C}$
$\Rightarrow \frac{B C - D C}{D C} = \frac{m}{l}$
$\Rightarrow \frac{K - D C}{D C} = \frac{m}{l}$
$\Rightarrow K l - D C . l = D C . m$
$\Rightarrow D C = \frac{K l}{m + l}$
Hence, A is the correct option.

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Δ A B C

l = log \frac{a^{2}}{b c}

m = log \frac{b^{2}}{c a}

n = log \frac{c^{2}}{a b}

l + m + n

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If $Δ A B C$ is isosceles with AB = AC and C(O,r) is the incircle of the $Δ A B C$ touching BC at L, prove that L bisects BC.

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