Question

Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the center.

Answer 1

Hint: Line joining the parallel tangents will be straight line and will pass through the centre as well. Choose the triangles consisting of angles subtending at centre by the intercept of tangent and prove them congruent to each other. Draw a neat diagram to observe those triangles. Use the RHS property to prove the angles to be congruent. Angle by a straight line is $${180}^{\circ }$$.

Complete step-by-step solution -
Here, it is given that the tangents drawn are parallel and hence, we need to prove that the intercept of the tangent between two parallel tangents to a circle subtends at a right angle at centre.

where OA, OB and OP are the radius of the circle and AM, BM, MN are parts of the tangents.

Now, it is given that the tangents $T_1$ and $T_2$ are parallel to each other.
It means line AB will be a straight line and angle AOB will be ${180}^{\circ }$ as per the property of parallel tangents in circle (proved in note part).
So, we get
$\mathrm{\angle }AOB={180}^{\circ }..........(i)$
Now, we know the angle made at the touching point of the tangent to the circle, with the tangent and radius is ${90}^{\circ }$ , i.e. tangent and radius will be perpendicular to each other. So, we can write relations with tangents $T_1,T_2,$ MN and radius OA, OB and OP respectively as:
$\begin{array}{c}\mathrm{\angle }OAM={90}^{\circ }\\ \mathrm{\angle }OBN={90}^{\circ }\\ \mathrm{\angle }OPN={90}^{\circ }\\ \mathrm{\angle }OPM={90}^{\circ }\end{array}].........(iii)$
Now, in $\mathrm{\Delta }AOM$ and $\mathrm{\Delta }PON,$ we have
$\mathrm{\angle }OAM=\mathrm{\angle }OPM={90}^{\circ }$ [from equation (iii)]
OM=OM (common)
OA = OP = Radius (Radius of circle)
Hence, $\mathrm{\Delta }AOM$ and $\mathrm{\Delta }PON$ are congruent to each other by RHS property. Hence, we get
$\mathrm{\angle }AOM=\mathrm{\angle }MOP$ (By C.P.C.T)………(iv)
Similarly, we can prove angle NOP and angle BON as equal angles using the same approach.
As,
$\mathrm{\angle }OBN=\mathrm{\angle }OPN$ [from equation (iii)]
ON = ON (common)
OP=OB (radius)
Hence, $\mathrm{\Delta }BON\cong \mathrm{\Delta }PON$. So, we get
$\mathrm{\angle }BON=\mathrm{\angle }NOP$ (By C.P.C.T)…………..(v)
Now, we can observe the angle AOB which can be written as sum of angles $\mathrm{\angle }AOM,\mathrm{\angle }MOP,\mathrm{\angle }NOP,\mathrm{\angle }BON$ in following way:
$\mathrm{\angle }AOB=\mathrm{\angle }AOM+\mathrm{\angle }MOP+\mathrm{\angle }NOP+\mathrm{\angle }BON$
Now, put $\mathrm{\angle }AOB={180}^{\circ }$ from the equation (i) to the above expression. So we get
$\mathrm{\angle }AOM+\mathrm{\angle }MOP+\mathrm{\angle }NOP+\mathrm{\angle }BON={180}^{\circ }..........(vi)$
Now, we can replace $\mathrm{\angle }MOP$ from equation (iv) and also can replace $\mathrm{\angle }BON$ by $\mathrm{\angle }NOP$ from equation (v). So, we get equation (vi) as
$\mathrm{\angle }MOP+\mathrm{\angle }MOP+\mathrm{\angle }NOP+\mathrm{\angle }NOP={180}^{\circ }$
$\Rightarrow 2\mathrm{\angle }MOP+2\mathrm{\angle }NOP={180}^{\circ }$
$$\Rightarrow 2(\mathrm{\angle }MOP+\mathrm{\angle }NOP)={180}^{\circ }$$
$\Rightarrow \mathrm{\angle }MOP+\mathrm{\angle }NOP={\displaystyle \frac{{180}^{\circ }}{2}}={90}^{\circ }$
Now, we can replace $\mathrm{\angle }MOP+\mathrm{\angle }NOP$ by $\mathrm{\angle }MON$ as both will be equal from the diagram. Hence, we get
$\mathrm{\angle }MON={90}^{\circ }$
So, the angle subtended by the tangent MN is ${90}^{\circ }$.
Hence, the given statement is proved

Note: One may confuse with the statement that the line AOB will be straight and from angle as ${180}^{\circ }$

Proof:
Given: $T_1$ is parallel to $T_2$ and O is the centre of the circle
$\begin{array}{rl}& \mathrm{\angle }OM{T}_1={90}^{\circ }\\ & \mathrm{\angle }ON{T}_2={90}^{\circ }\end{array}$
Construction: Draw a line passing through centre O, and parallel to both the tangents $T_1$ and $T_2$
Proof:
As we know, angles made by transversal on the same side of two parallel lines has sum of ${180}^{\circ }$. So, OM is a transversal and tangent $T_{!}$ and line T are parallel to each other. So, we get
$\mathrm{\angle }MOK+\mathrm{\angle }OM{T}_1={180}^{\circ }$
$\begin{array}{rl}& \mathrm{\angle }MOK+{90}^{\circ }={180}^{\circ }\\ & \mathrm{\angle }MOK={90}^{\circ }\end{array}$
Similarly, $\mathrm{\angle }NOK={90}^{\circ }$
So, $\mathrm{\angle }MON=\mathrm{\angle }MOK+\mathrm{\angle }NOK={90}^{\circ }+{90}^{\circ }={180}^{\circ }$
RHS criteria means one angle of both the triangles should be ${90}^{\circ }$, the hypotenuse side of both should be equal and one another side also be equal. So, don’t confuse with criteria applied for proving two triangles as congruent triangles.