Question

in the given figure ,AB=AC. If BO and CO The bisector of angle B and angle C meet at O and BC is produced to D prove that angle BOC = angle ACD

Answer 1

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Answer 2

Proved below.

Step-by-step explanation:

Given:

Here, AB = AC

BO and CO are the bisector of angle B and angle C meet at O and BC is produced to D.

In Δ ABC,

AB = AC              [ Given in figure]

∠ ACB = ∠ ABC         [ Given ]

$\frac{1}{2} ACB =\frac{1}{2} ABC$                   [ multipying both sides by $\frac{1}{2}$]

∠ MCB = ∠ MBC         [ Given ]      [1]

In Δ BMC,

∠ BMC + ∠ MBC + ∠ MCB = 180°          [ Measure of a triangle is equal to 180]

∠ BMC + 2 ∠ MCB = 180°                  [ from 1 ]

∠ BMC + ∠ ACB = 180°                       [2]

And , ∠ ACB + ∠ ACD = 180°   [ Sum of linear pair angles is 180° ]   [3]

From eq 2 and 3,

∠ ACB + ∠ BMC = ∠ ACB + ∠ ACD

∠ BMC = ∠ ACD

Hence proved.