Ex.10.6 Q4 Circles Solution - NCERT Maths Class 9

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Question

Let the vertex of an angle \(\begin{align}{ABC}\end{align}\) be located outside a circle and let the sides of the angle intersect equal chords \(\begin{align}{AD}\end{align}\) and \(\begin{align}{CE}\end{align}\) with the circle. Prove that \(\begin{align} \angle {ABC}\end{align}\) is equal to half the difference of the angles subtended by the chords \(\begin{align}{AC}\end{align}\) and \(\begin{align}{DE}\end{align}\) at the centre.

 Video Solution
Circles
Ex 10.6 | Question 4

Text Solution

What is known?

Point \(B\) lies outside the circle also \(AD\) and \(CE \) are equal chords. 

What is unknown?

To prove that \(\begin{align} \angle {ABC}\end{align}\) is equal to half the difference of the angles subtended by the chords \(\begin{align} {AC}\end{align}\) and \(\begin{align}{DE}\end{align}\) at the centre means \(\begin{align}\angle ABC = \frac{1}{2}\left( {\angle DOE - \angle AOC} \right)\end{align}\)

Reasoning:

  • The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
  • A quadrilateral \(\begin{align}{ACED}\end{align}\) is called cyclic if all the four vertices of it lie on a circle.
  • The sum of either pair of opposite angles of a cyclic quadrilateral is \( 180^\circ.\)
  • Using Side-Side-Side (SSS criteria) and Corresponding parts of congruent triangles (CPCT) we prove the statement.

Steps:

Consider \( \Delta {AOD}\) and \( \Delta {COE,}\)

\(\begin{align}{OA}&={OC} \qquad {\text { (Radii of the circle) }} \\{OD}&={OE} \qquad {\text { (Radii of the circle) }} \\ {AD}&={CE} \qquad {\text { (Given) }}\end{align}\)

\( ∴ \Delta {AOD} \cong \Delta {COE} \) (SSS congruence rule )

\(\angle {OAD}=\angle {OCE} \) (By CPCT )  ....(\(1\)

\( \angle {ODA}=\angle {OEC} \) (By CPCT )  .....(\(2\)

Also,

\(\!\angle {OAD}\!=\!\angle {ODA} \)(As \(OA=OD\))...(\(3\))

From Equations (\(1\)), (\(2\)), and (\(3 \)), we obtain

\[ \!\! \angle OAD \! = \!\angle OCE = \!\!\angle {ODA}= \!\! \angle {OEC} \]

\[\begin {align} \angle {OAD} & =\angle {OCE} \\ &=\angle {ODA}\\ &=\angle {OEC}\\ &=x \end {align}\]

In \(\begin {align} \Delta {OAC,} \end {align}\)

\[\begin{align} {OA} &={OC} \\ ∴ {OCA} &=\angle {OAC}(\text { Let } a) \end{align}\]

In \(\begin {align} \Delta {ODE,} \end {align}\)

\[\begin{align} \rm{OD} &=\rm{OE} \\ \angle \rm{OED} &=\angle \rm{ODE}\text{ (Let $y$)} \end{align}\]

\(\begin{align}\rm{ADEC}\end{align}\) is a cyclic quadrilateral.

\(∴ \angle {CAD}+\angle {DEC}=180^{\circ} \) (Opposite angles are supplementary)

\(\begin{align} x+a+x+y&=180^{\circ} \\ \!2 x+a+y&=180^{\circ} \\ y \!&=\!180^{\circ}-2 x \! -a ...\! (4) \end{align}\)

However, \(\angle {DOE}=180^{\circ}-2 {y}\)

And, \(\begin {align}\angle{AOC}=180^{\circ}-2 {a} \end {align}\)

\[\begin{align} & \angle {DOE}-\angle {AOC} \\&=2a-2y \\ & =2 a-2\left(180^{\circ}-2 x-a\right) \\ &=4 a+4 x-360^{\circ} \ldots \ldots  (5) \end{align}\]

\( \angle {BAC}+\angle {CAD}=180^{\circ} \)(Linear pair )

\(\begin{align}   ∴ \angle {BAC}&=180^{\circ}-\angle{CAD} \\ &=180^{\circ}-(a+x)\end{align}\)

Similarly, \(\begin {align} \angle {ACB}=180^{\circ}-(a+x) \end {align}\)

In \(\begin {align} \Delta {ABC,} \end {align}\)

\(\angle {ABC}+\angle {BAC}+\angle {ACB}=180^{\circ}\)

 (Angle sum property of a triangle)

\(\begin {align}\angle {ABC} &=180^{\circ}-\angle {BAC}-\angle {ACB} \\ &= \begin{Bmatrix} 180^{\circ} \\ -\left(180^{\circ}-a-x\right) \\ -\left(180^{\circ}-a-x\right) \end{Bmatrix} \\ &=2 a+2 x-180^{\circ} \\ &=\frac{1}{2}\left[4 a+4 x-360^{\circ}\right]\end {align}\)

Using Equation (\(5\))

\[\begin {align}\angle {ABC}=\frac{1}{2}[\angle {DOE}-\angle {AOC}] \end {align}\]

Hence it is proved that \(\begin{align} \angle {ABC}\end{align}\) is equal to half the difference of the angles subtended by the chords \(\begin{align}{AC}\end{align}\) and \(\begin{align}{DE}\end{align}\) at the centre.

  
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