Question 76

If I is the in centre of ABC and BIC = $$135^\circ$$, then ABC is

Solution

I is the incentre of a triangle

internal bisector of angle of a triangle meet at I

$$ < $$BIC = $$ 135^\circ $$ (given) ................(1)

$$ < $$ BIC = 90 + $$ \frac{1}{2} $$ <A .......................(2) (property of internal bisector)

from (1) and (2)

$$ 135^\circ = 90 + \frac{1}{2} <A $$

$$ 45^\circ = \frac{1}{2} <A $$

$$ <A = 90^\circ $$

therefore triangle ABC is a right angled triangle

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