Question

The point A, B, C are \[{{\rm{z}}_1}{\rm{,}}{{\rm{z}}_2}{\rm{,}}{{\rm{z}}_3}\]on the circumference of the circle drawn on OA as diameter, O being the origin. If \[\angle {\rm{ABC = }}\angle {\rm{BOC}}\] then prove that \[{{\rm{z}}_2}^2\cos 2{\rm{\theta = }}{{\rm{z}}_1}{{\rm{z}}_3}{\cos ^2}{\rm{\theta }}\].
Passage: Angle subtended by chord of a circle at the centre is twice the angle subtended at the circumference. If OP is rotated through an angle \[\phi \] an anticlockwise direction to become OQ. Then \[{\rm{OQ = OP}}{{\rm{e}}^{{\rm{i}}\phi }}\]

Answer 1

Hint:
Here we have to use the concept of the rotation of the line segment by an angle. So we have to make two equation by using this concept which is showing the relation between \[{{\rm{z}}_1}{\rm{,}}{{\rm{z}}_2}\] and \[{{\rm{z}}_1}{\rm{,}}{{\rm{z}}_3}\]. Then we have to write their relation in terms of the angle between them. Then by simply solving the equation we will get the required relation.

Complete step by step solution:
It is given that\[\angle {\rm{ABC = }}\angle {\rm{BOC}}\].
Now we have to use the concept of the rotation of a unit length line segment by an angle as given in the passage part of the question. So, we get
\[ \Rightarrow \dfrac{{{{\rm{z}}_2}}}{{\left| {{{\rm{z}}_2}} \right|}} = \dfrac{{{{\rm{z}}_1}}}{{\left| {{{\rm{z}}_1}} \right|}}{{\rm{e}}^{{\rm{i\theta }}}}\]………… (1)
\[ \Rightarrow \dfrac{{{{\rm{z}}_3}}}{{\left| {{{\rm{z}}_3}} \right|}} = \dfrac{{{{\rm{z}}_1}}}{{\left| {{{\rm{z}}_1}} \right|}}{{\rm{e}}^{{\rm{i\theta }}}}\]………… (2)
According to the question, the angle at B and C is \[{\rm{9}}{{\rm{0}}^0}\] and OA is the diameter. Therefore we get
\[ \Rightarrow \dfrac{{{\rm{OC}}}}{{{\rm{OA}}}} = \dfrac{{{{\rm{z}}_3}}}{{\left| {{{\rm{z}}_1}} \right|}} = \cos 2{\rm{\theta }}\]………… (3)
\[ \Rightarrow \dfrac{{{\rm{OB}}}}{{{\rm{OA}}}} = \dfrac{{{{\rm{z}}_2}}}{{\left| {{{\rm{z}}_1}} \right|}} = \cos {\rm{\theta }}\]………… (4)
Now we have to solve these equations and then we will get the required equation.
Now squaring equation (1) on both sides, we get
\[ \Rightarrow {\left( {\dfrac{{{{\rm{z}}_2}}}{{\left| {{{\rm{z}}_2}} \right|}}} \right)^2} = {\left( {\dfrac{{{{\rm{z}}_1}}}{{\left| {{{\rm{z}}_1}} \right|}}{{\rm{e}}^{{\rm{i\theta }}}}} \right)^2}\]
\[ \Rightarrow \dfrac{{{{\rm{z}}_2}^2}}{{{{\left| {{{\rm{z}}_2}} \right|}^2}}} = \dfrac{{{{\rm{z}}_1}^2}}{{{{\left| {{{\rm{z}}_1}} \right|}^2}}}{{\rm{e}}^{{\rm{2i\theta }}}}\]
Now by simplifying the above equation we get
\[ \Rightarrow {{\rm{z}}_2}^2 = \dfrac{{{{\rm{z}}_1}^2 \times {{\left| {{{\rm{z}}_2}} \right|}^2}}}{{{{\left| {{{\rm{z}}_1}} \right|}^2}}}{{\rm{e}}^{{\rm{2i\theta }}}}\]
Now by using the equation (4) in the above equation, we get
\[ \Rightarrow {{\rm{z}}_2}^2 = {\cos ^2}{\rm{\theta }}{{\rm{z}}_1}^2{{\rm{e}}^{{\rm{2i\theta }}}}\]…………….. (5)
Now by using the equation (2) and equation (3), we get
\[ \Rightarrow {{\rm{z}}_3} = \cos 2{\rm{\theta }}{{\rm{z}}_1}{{\rm{e}}^{{\rm{2i\theta }}}}\]…………….. (6)
Now we have to divide the equation (5) by equation (6), we get
\[ \Rightarrow \dfrac{{{{\rm{z}}_2}^2}}{{{{\rm{z}}_3}}} = \dfrac{{{{\cos }^2}{\rm{\theta }}}}{{\cos 2{\rm{\theta }}}}\]
\[ \Rightarrow {{\rm{z}}_2}^2\cos 2{\rm{\theta = }}{{\rm{z}}_1}{{\rm{z}}_3}{\cos ^2}{\rm{\theta }}\]
Hence proved

Note:
Chord of a circle is the line segment whose endpoint always lies on the circumference of the circle and it should be noted that the diameter of the circle is the longest chord which passes through the centre of the circle or we can say that the chords passing through the center of the circle is always the longest chord of that circle. Also radius of the circle is not known as the chord as one end point lies on the center of the circle. We should know that when a point lies on a curve or a line then that point satisfies the equation of that curve or line.