Question

In the given figure, BI is the bisector of ∠ABC and CI is the bisector of ∠ACB. Find ∠BIC

Answer 1

Step1: Find ∠ABC and ∠ACB

In ∆ABC, AB=AC

[ From the figure]

∴ ∠B = ∠C

[∵ Angles opposite to equal sides]

And ∠A + ∠B + ∠C = ${180}^{\circ }$

[∵ Angle sum property of a triangle]

${40}^{\circ }$ + ∠B + ∠B = ${180}^{\circ }$

2∠B = ${180}^{\circ }$ - ${40}^{\circ }$

2∠B = ${140}^{\circ }$

∴ ∠B = ${70}^{\circ }$

∴ ∠ABC = ∠ACB = ${70}^{\circ }$

Step2: Find ∠IBC and ∠ICB

Now, BI and CI are the bisector of ∠ABC and ∠ACB, respectively.

∴ ∠IBC = $\frac{1}{2}$∠ABC = $\frac{1}{2}$ × ${70}^{\circ }$

∠IBC = ${35}^{\circ }$

Similarly, ∠ICB = ${35}^{\circ }$

Step 3: Find ∠BIC using angle sum property of a triangle

In ∆IBC,

∠BIC + ∠IBC + ∠ICB = ${180}^{\circ }$

[∵ Angle sum property of a triangle]

∠BIC + ${35}^{\circ }$ + ${35}^{\circ }$ = ${180}^{\circ }$

∠BIC + ${70}^{\circ }$ = ${180}^{\circ }$

∠BIC = ${180}^{\circ }$ - ${70}^{\circ }$

= ${110}^{\circ }$

Hence, the value of ∠BIC is ${110}^{\circ }$.