Question
The common tangents AB and CD to two circles with centres O and O’ intersect at E between their centres. Prove that the points O, E and O’ are collinear.
✓
Answer
Joint AO, OC and O’D, O’B
Now, in $\triangle\text{EO}'\text{D}$ and,
O’D = O’B [radius]
O’E = O’E [common side]
ED = EB [since, tangentdrawn from an external point to the circle are equal in the length.]
$\therefore\triangle\text{EO'D}\cong\triangle\text{EO'B}$ [by SSS congruence rule]
$\angle\text{EO'D}=\angle\text{EO'B}$
O'E is the angle bisects of $\angle\text{DEB}\ ...\text{(i)}$
Similarily, OE is the angle bisects of $\angle\text{AEC}$
Now, in quadilateral DEBO'
$\angle\text{OD'E}=\angle\text{O'BE}=90^{\circ}$[since, CED is the tangent to the circle and O'D is the radius, i.e,$\text{O'D}\bot\text{CED}$]
$\Rightarrow\angle\text{O'DE}+\angle\text{O'BE}=180^{\circ}$
$\therefore\angle\text{DEB}+\angle\text{DO'B}=180^{\circ}\ ...\text{(ii)}$
[since, DEBo' is cyclic quadilateral]
since, AB is a straight line.
$\therefore\angle\text{AED}+\text{DEB}=180^{\circ}$
$\Rightarrow\angle\text{AED}+180^{\circ}-\angle\text{DO'B}=180^{\circ}$
Now, in $\triangle\text{EO}'\text{D}$ and,
O’D = O’B [radius]
O’E = O’E [common side]
ED = EB [since, tangentdrawn from an external point to the circle are equal in the length.]
$\therefore\triangle\text{EO'D}\cong\triangle\text{EO'B}$ [by SSS congruence rule]
$\angle\text{EO'D}=\angle\text{EO'B}$
O'E is the angle bisects of $\angle\text{DEB}\ ...\text{(i)}$
Similarily, OE is the angle bisects of $\angle\text{AEC}$
Now, in quadilateral DEBO'
$\angle\text{OD'E}=\angle\text{O'BE}=90^{\circ}$[since, CED is the tangent to the circle and O'D is the radius, i.e,$\text{O'D}\bot\text{CED}$]
$\Rightarrow\angle\text{O'DE}+\angle\text{O'BE}=180^{\circ}$
$\therefore\angle\text{DEB}+\angle\text{DO'B}=180^{\circ}\ ...\text{(ii)}$
[since, DEBo' is cyclic quadilateral]
since, AB is a straight line.
$\therefore\angle\text{AED}+\text{DEB}=180^{\circ}$
$\Rightarrow\angle\text{AED}+180^{\circ}-\angle\text{DO'B}=180^{\circ}$
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