Question

The sides \[\text{AB}\] and \[\text{AC}\] of a \[\vartriangle \text{ABC}\] are produced to \[\text{P}\] and \[\text{Q}\] respectively. If the bisector of the angles \[\angle \text{PBC}\] and \[\angle \text{QCB}\] intersect at \[\text{O}\], then show that \[\angle \text{BOC}={{90}^{\circ }}-\dfrac{1}{2}\angle \text{A}\].

Answer 1

Hint: In this question, we will use fact that the angle bisector of \[\angle \text{PBC}\] and \[\angle \text{QCB}\] divides the angle into two equal parts. Now using the exterior angle theorem for triangle which states that if a side of a triangle is extended, then the resultant angle would be equal to the sum of the two opposite interior angles of the given triangle. Thus we will get \[\angle \text{PBC}=\angle \text{A}+\angle \text{ACB}\] and \[\angle \text{QCB}=\angle \text{A}+\angle \text{ABC}\]. Then we will use the angle sum property of the triangle \[\vartriangle \text{BOC}\] and then do some substitutions to get the desired result.

Complete step-by-step answer:
We are given a triangle \[\vartriangle \text{ABC}\].
The side \[\text{AB}\] of \[\vartriangle \text{ABC}\] is produced to \[\text{P}\] and the side \[\text{AC}\] of \[\vartriangle \text{ABC}\] is produced to \[\text{Q}\] as shown in the figure given below.

Let us suppose \[\text{BO}\] bisects \[\angle \text{PBC}\].
Thus we have
\[\angle \text{OBC}=\dfrac{1}{2}\angle \text{PBC}....................\text{(1)}\]
Again since \[\text{CO}\] bisects \[\angle \text{QCB}\] we have
\[\angle \text{OCB}=\dfrac{1}{2}\angle \text{QCB}............................\text{(2)}\]
Now the Exterior angle theorem states that if a side of a triangle is extended, then the resultant angle would be equal to the sum of the two opposite interior angles of the given triangle.
Using this property, we get that
\[\angle \text{PBC}=\angle \text{A}+\angle \text{ACB}.............................\text{(3)}\]
and
\[\angle \text{QCB}=\angle \text{A}+\angle \text{ABC}................\text{(4)}\]
Now using the angle sum property of a triangle in \[\vartriangle \text{BOC}\] which states that the sum of the three interior angles of a triangle if equal to \[{{180}^{\circ }}\], we get
\[\angle \text{BOC}+\angle \text{OBC}+\angle \text{OCB}={{180}^{\circ }}\]
\[\Rightarrow \angle \text{BOC}={{180}^{\circ }}-\left( \angle \text{OBC+}\angle \text{OCB} \right).................(5)\]
Now using equation (1) and equation (2), we get
\[\begin{align}
  & \angle \text{OBC}=\dfrac{1}{2}\angle \text{PBC} \\
 & \text{=}\dfrac{1}{2}\left( \angle \text{A}+\angle \text{ACB} \right)............(6)
\end{align}\]
Now using equation (2) and equation (3), we get
\[\begin{align}
  & \angle \text{OCB}=\dfrac{1}{2}\angle \text{QCB} \\
 & \text{=}\dfrac{1}{2}\left( \angle \text{A}+\angle \text{ABC} \right).............(7)
\end{align}\]
We will now substitute the values of equation (6) and equation (7) in equation (5).then we will get
\[\begin{align}
  & \angle \text{BOC}={{180}^{\circ }}-\left( \angle \text{OBC+}\angle \text{OCB} \right) \\
 & ={{180}^{\circ }}-\left( \dfrac{1}{2}\left( \angle \text{A}+\angle \text{ACB} \right)\text{+}\dfrac{1}{2}\left( \angle \text{A}+\angle \text{ABC} \right) \right) \\
 & ={{180}^{\circ }}-\left( \dfrac{1}{2}\left( \angle \text{A} \right)+\dfrac{1}{2}\left( \angle \text{A}+\angle \text{ABC}+\angle \text{ACB} \right) \right)
\end{align}\]
Here in the above equation, we have the sum of the angle of \[\vartriangle \text{ABC}\] is \[{{180}^{\circ }}\] using the angle sum property.
Thus we get
\[\begin{align}
  & \angle \text{BOC}={{180}^{\circ }}-\left( \dfrac{1}{2}\left( \angle \text{A} \right)+\dfrac{1}{2}\left( \angle \text{A}+\angle \text{ABC}+\angle \text{ACB} \right) \right) \\
 & ={{180}^{\circ }}-\left( \dfrac{1}{2}\left( \angle \text{A} \right)+\dfrac{1}{2}\left( {{180}^{\circ }} \right) \right) \\
 & ={{180}^{\circ }}-\dfrac{1}{2}\angle \text{A}-\dfrac{{{180}^{\circ }}}{2} \\
 & ={{180}^{\circ }}-{{90}^{\circ }}-\dfrac{1}{2}\angle \text{A} \\
 & \text{=9}{{0}^{\circ }}-\dfrac{1}{2}\angle \text{A}
\end{align}\]
Hence the statement is proved.

Note: In this problem, trivial calculation mistakes can be made. So take care of that. Use the exterior angle property of triangles carefully while choosing the sum of the interior angles of the triangle.