If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that angleDBC = 120 degree, prove that BC + BD = BO, i.e., BO = 2BC.

Name: If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that angleDBC = 120 degree, prove that BC + BD = BO, i.e., BO = 2BC.
Uploaded: Sat, 2021-01-30 10:14
Description: If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that angleDBC = 120 degree, prove that BC + BD = BO, i.e., BO = 2BC.

#Circles
#Maths
#Exemplar Maths for Class 10
#NCERT

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If from an external point B of a circle with centre O, two tangents BC andBD are drawn such that $<$ DBC = $120^{\circ}$ , prove that BC + BD = BO, i.e., BO = 2BC.

Answers (1)

Solution
According to question

Given : $\angle DBC= 120^{\circ}$
To Prove : BC + BD = BO, i.e., BO = 2BC
Here OC $\perp$ BC, OD $\perp$ BD ( $\because$ BC, BD are tangents)
Join OB which bisect $\angle DBC$
In $\bigtriangleup ODB$
$\cos \theta = \frac{B}{H}$
$\cos 60^{\circ}= \frac{BD}{OB}$
$\frac{1}{2}= \frac{BD}{OB}$
$OB= 2BD$
$OB= 2BC$ $\left ( \because BD= BC \, \ \ Tangents \right )$
$OB= BC+BC$
$OB= BC+BD\; \; \left ( \because BC= BD \right )$
Hence Proved