Problem (1)
Given : $ \displaystyle SPT$ is the tangent to the circle at $ \displaystyle P$.
$ \displaystyle PQ$ and $ \displaystyle RQ$ are the chords of the circle.
$ \displaystyle PM\bot RQ$ and $ \displaystyle RN\bot SPT$.
Prove : $ \displaystyle MN\parallel PQ$
Show/Hide Solution
$ \displaystyle \begin{array}{l}\ \ \ \ \ \angle QPT=\angle QRP\ (\angle \ \text{between tangent and chord }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{= }\angle \ \text{in alternate segment})\\\\\ \ \ \ \ \text{Since}\ PM\bot RQ\ \text{and}\ RN\bot SPT,\ \\\\\ \ \ \ \ \angle PMR=\angle PNR=90{}^\circ \\\\\therefore \ \ \ \angle PMR+\angle PNR=180{}^\circ \\\\\therefore \ \ \ PMRN\ \text{is cyclic}\text{.}\\\\\therefore \ \ \ \angle PNM=\angle QRP\ \ (\angle \ \text{in same arc)}\\\\\therefore \ \ \ \angle PNM=\angle QPT\\\\\ \ \ \ \ \text{Since}\ \angle PNM\ \text{and}\ \angle QPT\ \text{are alternating angles,}\\\\\ \ \ \ \ MN\parallel PQ.\ \end{array}$
Problem (2)
In the fgure, $ \displaystyle AP$ is a tangent to the circle at $ \displaystyle A$ and $ \displaystyle AP$ is parallel to $ \displaystyle BQ$. Prove that
(i) $ \displaystyle ∆ABC$ is similar to $ \displaystyle ∆AQB,$
(ii) $ \displaystyle AB^2 = AQ × AC.$
Show/Hide Solution
$ \displaystyle \begin{array}{l}\ \ \ \ \ \alpha =\beta \ (\angle \ \text{between tangent and chord }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{= }\angle \ \text{in alternate segment})\\\\\ \ \ \ \ \text{Since}\ AP\parallel BQ,\ \ \alpha =\theta .\\\\\ \ \ \ \ \text{In }\Delta ABC\ \text{and}\ \Delta AQB,\\\\\ \ \ \ \ \alpha =\theta \ \text{(proved)}\\\\\ \ \ \ \ \delta =\delta \ \ \text{(common }\angle \text{)}\\\\\therefore \ \ \ \Delta ABC\sim \Delta AQB\ \ (\text{AA corollary)}\end{array}$
$ \displaystyle \therefore \ \ \ \frac{{AB}}{{AQ}}=\frac{{AC}}{{AB}}$
$ \displaystyle \therefore \ \ \ A{{B}^{2}}=AQ\times AC$
Problem (3)
In the fgure, $ \displaystyle O$ is the centre of the circle, $ \displaystyle PQ$ is a diameter and $ \displaystyle AB$ is a chord which is parallel to $ \displaystyle PQ.$ $ \displaystyle AQ$ and $ \displaystyle OB$ intersect at $ \displaystyle X$. Prove that $ \displaystyle ∠BXQ = 3∠PQA$.
Show/Hide Solution
$ \displaystyle \begin{array}{l}\ \ \ \ \text{Since}\ AB\parallel PQ,\\\\\ \ \ \ \beta =\phi .\ \ \ \text{(}\because \text{alternating }\angle \text{)}\\\\\ \ \ \ \alpha =\theta .\ \ \ \text{(}\because \text{alternating }\angle \text{)}\end{array}$
$ \displaystyle \ \ \ \ \alpha =\frac{1}{2}\phi \ \ \ \text{(}\because \text{inscribed }\angle =\frac{1}{2}\text{central }\angle \text{)}$
$ \displaystyle \begin{array}{l}\therefore \ \ \phi =\beta =2\alpha =2\theta \\\\\ \ \ \ \text{In }\Delta ABX,\ \\\\\ \ \ \ \gamma =\beta +\alpha \\\\\ \ \ \ \ \ \ =2\alpha +\alpha \\\\\ \ \ \ \ \ \ =3\alpha \\\\\ \ \ \ \ \ \ =3\theta \ \ \ \ (\because \alpha =\theta )\\\\\therefore \ \ \angle BXQ=3\angle PQA\end{array}$
Problem (4)
In the fgure, $ \displaystyle BD$ and $ \displaystyle CE$ are tangents to the circle, of which $ \displaystyle AB$ is a diameter and $ \displaystyle ACD$ is a straight line. Prove that
(i) $ \displaystyle \angle ABC=\angle ECD$
(ii) $ \displaystyle BE = ED.$
Show/Hide Solution
$ \displaystyle \begin{array}{l}\ \ \ \text{It is obvious that}\\\\\ \ \ \beta =\gamma \ \ (\because \angle \ \text{between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\angle \ \text{in alternate segment})\\\\\ \ \ \text{But, }\gamma =\theta \ \ (\because \text{vertically opposite }\angle \text{s)}\\\\\therefore \ \beta =\theta \\\\\therefore \angle ABC=\angle ECD\\\\\ \ \ \ \text{Since }AB\ \text{is a diameter, }\\\\\ \ \ \ \angle ACB=90{}^\circ \ \ (\because \angle \ \text{in semicircle)}\\\\\therefore \ \ \angle BCD=90{}^\circ \\\\\ \ \ \ \text{Furthermore, }AB\ \text{is a diameter and }BD\ \text{is a tangent,}\\\\\ \ \ \ AB\bot BD.\\\\\therefore \ \ \Delta ABD\sim \Delta ACB\sim \Delta BCD\\\\\therefore \ \ \beta =\delta \\\\\therefore \ \ \theta =\delta \\\\\therefore \ \ \Delta BCD\ \text{is an isosceles triangle with base }CD.\\\\\therefore \ \ EC=ED\\\\\ \ \ \ \text{Since}\ BE=EC,\ \ (\because \text{tangents from same exterior point)}\\\\\ \ \ \ BE=ED\end{array}$
Problem (5)
In the figure, $ \displaystyle ABCD$ and $ \displaystyle APRD$ are two circles intersecting at $ \displaystyle A$ and $ \displaystyle D$. $ \displaystyle ARC$ and $ \displaystyle BPD$ are straight lines. $ \displaystyle DR$ produced meets $ \displaystyle BC$ at $ \displaystyle S$. Prove that $ \displaystyle PR$ is parallel to $ \displaystyle BC.$
Show/Hide Solution
In circle $ \displaystyle APRD, \alpha = \phi \ \ \ (\angle \text{s}\ \text{in}\ \text{same}\ \text{arc})$
Similarly, In circle $ \displaystyle ABCD, \alpha = \beta \ \ \ (\angle \text{s}\ \text{in}\ \text{same}\ \text{arc})$
$ \displaystyle \therefore\ \phi= \beta$
Since $ \displaystyle \phi$ and $ \displaystyle \beta$ are corresponding angles,
$ \displaystyle PR \parallel BC$
Problem (6)
In the figure, $ \displaystyle HAE$ is the tangent to the circle at $ \displaystyle H, BH = BE$ and $ \displaystyle KH$ is the angle bisector of $ \displaystyle ∠BHE$ and it cuts the circle at $ \displaystyle D.$ Given that $ \displaystyle BD$ produced meets $ \displaystyle HE$ at $ \displaystyle A,$ prove that
(i) $ \displaystyle HD = BD,$
(ii) $ \displaystyle A, D, K$ and $ \displaystyle E$ are concyclic.
Show/Hide Solution
$ \displaystyle \begin{array}{l}\ \ \ \beta =\phi ,\ \ \ (\angle \ \text{between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\angle \ \text{in alternate segment})\end{array}$
$ \displaystyle \ \ \ \text{Since}\ KH\ \text{is}\ \text{the}\ \text{angle}\ \text{bisector}\ \text{of}\ ∠BHE,$
$ \displaystyle \ \ \ \theta = \phi$
$ \displaystyle \therefore \theta = \beta$
$ \displaystyle \therefore HD = BD$
$ \displaystyle \ \ \ \text{Since}\ BH = BE,$
$ \displaystyle \ \ \ \varepsilon=\theta + \phi$
$ \displaystyle \therefore \varepsilon=\theta + \beta$
$ \displaystyle \ \ \ \text{In}\ \triangle BDH,$
$ \displaystyle \ \ \ \alpha=\theta + \beta$
$ \displaystyle \therefore \varepsilon=\alpha$
$ \displaystyle \text{Since}\ \alpha +\delta =180{}^\circ ,$
$ \displaystyle \ \ \ \varepsilon +\delta =180{}^\circ ,$
$ \displaystyle \therefore \ A,D,K,E\ \text{are concyclic}\text{.}$
