Question

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Using above result , prove the following:
In figure , XZ touches the circle with centre O at Y. Diameter BA when produced meets XZ at X . Given $\angle BXY=b$ and $\angle AYX=a$, then prove that $b+2a={90}^{\circ }$

Answer 1

Consider the $\Delta OXY$,

Since, $\angle OXY=b$ and $\angle XYO={90}^{\circ }$

So,

$\angle OXY+\angle XYO+\angle YOX={180}^{\circ }$

$⟹b+{90}^{\circ }+\angle YOX={180}^{\circ }$

$⟹\angle YOX$ = ${90}^{\circ }$ - b

Since, $\angle YOX=\angle XOY=\angle AOY$

$⟹\angle AOY$ = ${90}^{\circ }$ - b

Since, OA and OY are the radius of $\Delta OAY$

So, $\angle OAY=\angle OYA$

So,

$\angle OAY+\angle OYA+\angle AOY={180}^{\circ }$

$⟹\angle AOY+2\angle OAY$= ${180}^{\circ }$

$⟹{90}^{\circ }-b+2\angle OAY$= ${180}^{\circ }$

$⟹2\angle OAY$= ${180}^{\circ }-{90}^{\circ }+b$

$⟹2\angle OAY$= ${90}^{\circ }$ + b

$⟹\angle OAY$ = $\frac{{90}^{\circ }+b}{2}$

Since, $\angle AYX=a$ and $\angle XYO={90}^{\circ }$

$⟹\angle OAY$= ${90}^{\circ }$ - a

So,

${90}^{\circ }$ - a = $\frac{{90}^{\circ }+b}{2}$

$⟹{90}^{\circ }+b=2\times ({90}^{\circ }-a)$

$⟹{90}^{\circ }+b={180}^{\circ }-2a$

$⟹b+2a={180}^{\circ }-{90}^{\circ }$

$⟹b+2a={90}^{\circ }$.

Hence, proved.