Chapter 9.4, Problem 23WE

To determine

To find: In given figure, if a line from center O, perpendicular to tangent, bisects its corresponding chord.

Answer to Problem 23WE

Given figure suggest the statement that if a perpendicular is drawn on point of contact of tangent on it, it bisects the chord of another concentric circle, so that tangent and chord coincide each other and is proved also.

Explanation of Solution

Given information: In below given figure, $AB$ is tangent to smaller circle but it is a chord on outer circle,

  

Concept used: (i) Any line from center to point of contact of tangent on it, is perpendicular on it.

(ii) Using HL congruency, in two right triangles, if corresponding hypotenuse and any one pair of another legs is congruent then two triangles are congruent.

Calculation: In given figure,

Given : $AC$ is tangent to smaller circle. Also, $AO=OC$ (Radii of outer circle)

To prove : $AB=BC$

Construction: Draw perpendicular $OB\perp AC$ so that $\angle OBA=\angle OBC={90}^0$ .

Proof: In right triangles $\Delta OBA$ and $\Delta OBC$ ,

  $\begin{array}{l}(i)OA=OC(\text{Radii)}\\ \text{(ii) }OB=OB\mathrm{Re}\text{flexive property}\end{array}$

So, by HL (Hypotenuse-leg) congruency, these two triangles are congruent. Hence, by CPCT (corresponding parts of congruent triangles) property, $AB=BC$

Conclusion: Thus, It proves that any perpendicular from the center of a circle to its chord, always bisects this chord.