Question

Bisectors of angles A, B and C of a triangle ABC intersects its circumcircle at D, E and F respectively. Prove that the angles of the triangles DEF are $90^\circ - \dfrac{1}{2}A$, $90^\circ - \dfrac{1}{2}B$ and $90^\circ - \dfrac{1}{2}C$.

Answer 1

Hint: In this problem, the property of angle addition is used. According to the angle addition property two adjacent angles can be added to determine the other unknown angles.

Complete step-by-step solution:Given:
The bisectors of angles A, B and C of a triangle ABC intersects its circumcircle at D, E and F respectively.
                                                     

The following is the schematic diagram of a triangle ABC inscribed in a circle.
In triangle $\Delta DEF$, we can say that
\[\angle D = \angle EDF\]
Also, by the angle addition property, we can write
\[\angle EDF = \angle EDA + \angle FDA\]
Now, the angles inscribed in the same arc are equal.
$\angle EDA = \angle EBA$ and $\angle FDA = \angle FCA$
Since, BE is bisector of \[\angle B\] and CF is bisector of \[\angle C\]. Now, the angle $\angle EDF$ will be equal to,
$\begin{array}{c}
\angle EDF = \angle EBA + \angle FCA\\
\angle EDF = \dfrac{1}{2}\angle B + \dfrac{1}{2}\angle C
\end{array}$
The above equation can be written as:
$\angle D = \dfrac{{\angle B + \angle C}}{2}$……(i)
Also, the angle $\angle E$ is equal to,
$\angle E = \dfrac{{\angle A + \angle C}}{2}$…..(ii)
Also, the angle $\angle F$ is equal to,
$\angle F = \dfrac{{\angle A + \angle B}}{2}$….(iii)
Now, we know that the sum of all the angles of a triangle is equal to $180^\circ $.
\[\begin{array}{c}
\Rightarrow\angle A + \angle B + \angle C = 180^\circ \\
\Rightarrow\angle B + \angle C = 180^\circ - \angle A
\end{array}\]….(iv)
$\Rightarrow$\[\angle A + \angle C = 180^\circ - \angle B\]………….(v)
$\Rightarrow$\[\angle A + \angle B = 180^\circ - \angle C\]……..…(vi)

On substituting the value from (iv) to equation (i),we get
\[\begin{array}{c}
\Rightarrow \angle D = \dfrac{{180^\circ - \angle A}}{2}\\
\Rightarrow\angle D = 90^\circ - \dfrac{{\angle A}}{2}
\end{array}\]
On substituting the value from (v) to equation (ii),we get
\[\begin{array}{c}
\Rightarrow\angle E = \dfrac{{180^\circ - \angle B}}{2}\\
\Rightarrow\angle E = 90^\circ - \dfrac{{\angle B}}{2}
\end{array}\]
On substituting the value from (vi) to equation (iii),we get
\[\begin{array}{c}
\Rightarrow\angle F = \dfrac{{180^\circ - \angle C}}{2}\\
\Rightarrow\angle F = 90^\circ - \dfrac{{\angle C}}{2}
\end{array}\]
Hence, it is proved that the angles of triangle DEF are \[\angle D = 90^\circ - \dfrac{{\angle A}}{2}\], $90^\circ - \dfrac{1}{2}\angle B$ and $90^\circ - \dfrac{1}{2}\angle C$.

Note: In this problem, it is important to know that BE is bisector of \[\angle B\] and CF is bisector of \[\angle C\]. We have also used the Basic property of a triangle, that the sum of all the angles of a triangle is equal to $180^\circ $.