Question

STATEMENT - 1: If tangents OR, PR, PQ and drawn respectively at A, B, C to the circle circumscribing an acute-angled $\Delta$ ABC so as the form another $\Delta$PQR, then the $\angle RPQ=\angle BAC$
STATEMENT - 2: $\Delta$ABC is similar to $\Delta$CPB.

• A. Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1
• B. Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1
• C. Statement - 1 is True, Statement - 2 is False
• D. Statement - 1 is False, Statement - 2 is True

Answer 1

The correct option is A Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

Given : An acute angled $△ABC,$ circumscribed by a circle and tangents $QR,PR$ and $PQ$ are drawn at $A,B$ and $C$ such that they form a $△PQR.$

Construct $OP,OB$ and $OC$ where $O$ is the centre of circle circumscribing $△ABC.$

$OB\perp PR$ and $OC\perp PQ.$

[The point of contact of radius and tangents to circle measure $90°.$]

In $△OBP$ and $△OCP,$

$OB=OC$ [Radic of same circle.]

$OP=OP$ [Common.]

$BP=CP$ [Length of tangents from an Extend point to circle are equal. ]

$△OBP\cong △OCP$ [$SSS$]

$\angle BPO=\angle CPO$ $\left[CPCT\right]$

Let $\angle BPO=\angle PO=a.$

In $△OBP,$

$\tan \alpha =\frac{OB}{BP}=\frac{r}{BP}.$

Where $a$ is radius of circle circumscribing $△ABC.$

Let $BP=x$

So, $\angle RPO=\tan 2\alpha$

$=\frac{2\tan \alpha }{1+{\tan }^2\alpha }=\frac{2\frac{r}{x}}{1+\frac{r^2}{x^2}}=\frac{2rx}{x^2+r^2}$

$\therefore$ $\angle ABC\cong \angle CPB$

$⟹$ Statement 1 is true & statement 2 is a correct explanation for statement 1.