
In the given figure, the common tangents $AB$ and $CD$ to two circles with centers $O$ and ${O}'$ intersect at E. Prove that the point $O, E\,and\,{E}'$ are collinear.
Answer
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Hint: To solve the above question, we will have to join \[OA\] and \[OC\] where \[AB\] and $CD$ are common tangents to circles with center $O$ and ${O}'$ intersecting at $E$. We have to show $O,\, E,\,{E}'$ are collinear. So for that, we have to know the concept of collinear that is three or more points that lie on the same line are collinear points. Example: The points $A,B\,and\,C$ lie on the line \[m\] . So, they are collinear. Here $OA$ and $OC$ are the radii of the same circle and $\angle OAE$&$\angle OCE$ are the right angle. We will consider the four rules to prove triangle congruence. They are called the $SSS$ rule, $SAS$ rule, $ASA$ rule, and $AAS$ rule
Complete step-by-step solution:
The figure for the given problem is as follows:
In triangle $OAE$ and triangle $OCE$, we have,
$OA=OC$ (Radii of same circle)
$OE=OE$ (Common)
$\angle OAE=\angle OCE$$\left( {{90}^{\circ}} \right)$ (As the tangent is always perpendicular to the radius at the point of contact)
$\therefore \Delta OAE\cong \,\Delta OCE$
So, we have $\angle AEO=\angle CEO$ (By$CPCT$ ) (1)
Similarly we have,$\angle DE{O}'=\angle BE{O}'$ (By$CPCT$ ) (2)
NOW $\angle AEC=\angle DEB$ (Vertically opposite angles)
$\Rightarrow \dfrac{\angle AEC}{2}=\dfrac{\angle DEB}{2}$
$\Rightarrow \angle AEO=\angle DE{O}'$
$\Rightarrow \angle AEO=\angle CEO=\angle DE{O}'=\angle BE{O}'$
So, we can see that all $4\,\angle 'S$ are equal and bisected by $OE$&$O{E}'$.
So, we can see that $O, E,{O}'$ are collinear.
Hence, in this way we have proved this problem.
Note: For to prove the following problem students have to know the properties of congruence of triangles. By using this property prove the following problem. Basically, students have to keep in mind that the concept that the tangent is always perpendicular to the radius at the point of contact and prove it accordingly. Sometimes, they do a mistake by taking the right angle. So, students have to be careful about it.
Complete step-by-step solution:
The figure for the given problem is as follows:
In triangle $OAE$ and triangle $OCE$, we have,
$OA=OC$ (Radii of same circle)
$OE=OE$ (Common)
$\angle OAE=\angle OCE$$\left( {{90}^{\circ}} \right)$ (As the tangent is always perpendicular to the radius at the point of contact)
$\therefore \Delta OAE\cong \,\Delta OCE$
So, we have $\angle AEO=\angle CEO$ (By$CPCT$ ) (1)
Similarly we have,$\angle DE{O}'=\angle BE{O}'$ (By$CPCT$ ) (2)
NOW $\angle AEC=\angle DEB$ (Vertically opposite angles)
$\Rightarrow \dfrac{\angle AEC}{2}=\dfrac{\angle DEB}{2}$
$\Rightarrow \angle AEO=\angle DE{O}'$
$\Rightarrow \angle AEO=\angle CEO=\angle DE{O}'=\angle BE{O}'$
So, we can see that all $4\,\angle 'S$ are equal and bisected by $OE$&$O{E}'$.
So, we can see that $O, E,{O}'$ are collinear.
Hence, in this way we have proved this problem.
Note: For to prove the following problem students have to know the properties of congruence of triangles. By using this property prove the following problem. Basically, students have to keep in mind that the concept that the tangent is always perpendicular to the radius at the point of contact and prove it accordingly. Sometimes, they do a mistake by taking the right angle. So, students have to be careful about it.
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