Chapter 9.4, Problem 24WE

To determine

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

Answer to Problem 24WE

Given figure suggests 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

Giveninformation: In below given figure,lis a tangent to the circle and $\overline{DE}$ is a chord in it.

  

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

(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 : lis tangent to given circle. Also, $OD=OE$ (Radii of outer circle)

To prove: $DA=EA$ , or chord is bisected by the perpendicular on it from center.

Construction: Draw perpendicular $OF\perp l$ so that $\angle OAD=\angle OFl={90}^0$ .

Proof: In right triangles $\Delta OAD$ and $\Delta OAE$ ,

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

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

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