Question

Bisectors of angle b and angle c intersect at o. prove that angle boc = 90 + 1/2 angle a

Answer 1

Answer:

Refer to the attachment for the answer.

Answer 2

Proof:

• we have to prove that, Δ$BOC=90^{0} +$Δ$\frac{A}{2}$.

• From Δ$ABC$, by using the angle sum property we can write as,

           Δ$A+$Δ$B+$Δ$C=180^{0}$        

• We know that Δ$B=2$Δ$CBO$ and Δ$C=2$Δ$OCB$, therefore sub in the original equation

        Δ$A$$+2$Δ$CBO+2$Δ$OCB =180^{0}$

        Δ$A$$+2($$CBO+OBC)$$=180^{0}$

hereby multiplying on both numerator and denominator with 2 for A we can get common as 2'

            Δ$\frac{A}{2} +$Δ$CBO+$Δ$OBC=\frac{180}{2}$

             Δ$CBO+$Δ$OBC=90^{0} -\frac{A}{2}$----->eq1

• From Δ$BOC$, by using the angle sum property we can write as,

       $CBO+OCB+BOC=180^{0}$

let's sub eq1 here, then we get

   $90^{0} -\frac{A}{2} +BOC=180^{0}$

∴ Δ$BOC=90^{0} +\frac{A}{2}$

Hence proved Δ$BOC=90^{0} +\frac{A}{2}$

#SPJ3