Question

If in a triangle ABC the external bisector of angle ABC and angle ACB meet at O then proof that angle BOC =90degree -1/2 angle BAC​

Answer 1

Answer:

Since bd is perpendicular it forms triangle abd with half angle b that is 45 degree

then consider triangle bdc we have at b as 45 degrees and d as 90 degree then by angle sum property of triangle

we have

90+45+angle c=180

135+c=180

c=180-135=45 degrees

hence angle abd =angle acb

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Answer 2

Step-by-step explanation:

The answers is in the picture