Ex.10.6 Q8 Circles Solution - NCERT Maths Class 9

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Question

Bisectors of angles \(\begin{align}  {A, B}\end{align}\) and \(\begin{align}  {C}\end{align}\) of a triangle \(\begin{align}  {ABC}\end{align}\) intersect its circumcircle at \(\begin{align} {D, E}\end{align}\) and \(\begin{align}{F}\end{align}\) respectively. Prove that the angles of the triangle \(\begin{align}  {DEF}\end{align}\) are \(\begin {align} 90^{0}-\frac{1}{2} {A}, \,90^{\circ}-\frac{1}{2} {B},\, 90^{\circ}-\frac{1}{2} {C}\end {align}\)

 Video Solution
Circles
Ex 10.6 | Question 8

Text Solution

What is known?

Bisectors of angles \(\begin {align} {A, B} \end {align}\) and \(\begin {align} {C} \end {align}\) of a triangle \(\begin {align} {ABC} \end {align}\) intersect its circumcircle at \(\begin {align} {D, E} \end {align}\) and \(\begin {align} {F } \end {align}\) respectively.

What is unknown?

Proof that the angles of the triangle \(\begin {align} \text{DEF} \end {align}\) are

\[ 90^{0}-\frac{1}{2} {A}, \;\; 90^{\circ}-\frac{1}{2} {B}, \;\; 90^{\circ}-\frac{1}{2} {C} \]

Reasoning:

Angles in the same segment are equal.

Steps:

It is given that \(\begin {align} {BE}\end {align}\) is the bisector of \(\begin {align} \angle {B.}\end {align}\)

\(\begin{align} \angle {ABE}=\frac{\angle {B}}{2} \end{align}\)

However,

\(\begin{align} \angle {ADE}&=\angle {ABE} \\\text { (Angles in} &\text{ the same}  \text{  segment }\\&  \text{ for chord $AE$ })\\ \angle {ADE}&=\frac{\angle {B}}{2}\end{align}\)

Similarly,

\(\begin{align}\angle {ADF}&=\angle {ACF}\\&=\frac{\angle {C}}{2}\\\text { (Angles in} & \text{ the same}  \text{  segment } \\ &\text{ for chord $AF$ })\end {align}\)

\[\begin{align} \angle {D} &=\angle {ADE}+\angle {ADF} \\ &=\frac{\angle {B}}{2}+\frac{\angle {C}}{2} \\ &=\frac{1}{2}(\angle {B}+\angle {C}) \\ &=\frac{1}{2}\left(180^{\circ}-\angle {A}\right) \end{align}\]

Similarly it can be proved for

\[\begin{align} \angle {E} &=\frac{1}{2}\left(180^{\circ}-\angle {B}\right) \\ \angle {F} &=\frac{1}{2}\left(180^{\circ}-\angle {C}\right) \end{align}\]

  
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