Question

in the given figure if bc = 20 cm and angle bad = angle cad find bd​

Answer 1

$\underline{\pink{ Angle \:Bisector \: Theorem: }}$

If a ray bisects an angle of a triangle, then it divides the opposite side of the triangle into segments that are proportional to the other two sides.

$Form \: figure \:(1)$

$\boxed{\blue{\frac{a}{b} = \frac{x}{y} }}$

$Here, In \: triangle \:ABC ,AB = 12 \:cm \\AC = 18 \:cm , BD = 20 \: cm \: and \: \angle {BAD} = \angle { CAD }$

$Let \: BD = x \:cm \: and \:DC = ( 20-x) \:cm$

$From \: figure \:(2) ,$

$\frac{AB}{AC} = \frac{BD}{DC}$

$\implies \frac{12}{18} = \frac{x}{20-x}$

$\implies \frac{2}{3} = \frac{x}{20-x}$

$\implies 2(20-x) = 3x$

$\implies 40 - 2x = 3x$

$\implies 40 = 3x + 2x$

$\implies 40 = 5x$

$\implies x = \frac{40}{5}$

$\implies x = 8 \:cm$

Therefore.,

$\red{ Value \:of \:BD } \green { = 8\:cm }$

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