Question

if I is the incentre of triangle ABC and angle BIC is 135° then what kind of triangle ∆ABC is?​

Answer 1

Answer:

hope it will help you...

Answer 2

$\angle \mathrm{BIC}=90^{\circ}+\angle \mathrm{A} / 2 }$$\angle \mathrm{BIC}=90^{\circ}+\angle \mathrm{A} / 2 }$Answer:

It is a Right angled triangle.

Step-by-step explanation:

In ΔABC,$\angle \mathrm{BPC}=90^{\circ}+\angle \mathrm{A} / 2 }$ [P is incentre of △ABC].

$\135^{\circ}=90^{\circ}+\angle \mathrm{A} / 2 \Rightarrow \angle \mathrm{A} / 2=45^{\circ} \Rightarrow \angle \mathrm{A}=90^{\circ}$

If one angle of a triangle is at 90°, then the triangle is a right angle triangle.

∴△ABC is right-angle triangle.

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