Chapter 9.7, Problem 28WE

To determine

To find:Measure of ED in given diagram.

Answer to Problem 28WE

The measure of $ED$ is 2.

Explanation of Solution

Giveninformation: In below given figure, $\overline{CD}$ is a tangent, arc $AC\cong \text{ arc }BC$ , $AB=3,AF=6,FE=10$ .

  

Concept used:Use the following theorems:

1. A chord subtends equal angles at a tangent as well as on its remaining arc.
2. If a secant and a tangent join at an exterior point, then the square of the tangent is equal to the product of secant and its exterior part of circle.
3. In two triangles if two pairs of corresponding angles are congruent, then these two triangles are similar.

Proof:Let $ED=x,CD=y$ .

Now angle formed by arc $AC$ on remaining arc is $\angle ABC$ and angle formed by arc BC will be equal to $\angle BCD$ by using (i) theorem.

So, in $\Delta ABF$ and $\Delta FCD$ , as arc $AC\cong \text{ arc }BC$ , so,

  $\angle ABC=\angle BCD\mathrm{......}(i)$

  $\angle AFB=\angle CFD(\text{Vertical }\angle \text{s)}\mathrm{......}(ii)$

So by AA similarity, both these triangles are similar, so the ratio of their corresponding sides will be equal, or,

  $\begin{array}{l}\frac{CD}{AB}=\frac{DF}{AF}\\ \Rightarrow \frac{y}{3}=\frac{10+x}{6}\\ \Rightarrow y=\frac{10+x}{2}\end{array}$

Again using (ii) theorem,

  $\begin{array}{l}C{D}^2=AD\times DF\\ y^2=(10+6+x)\times x\\ y^2=x(16+x)\end{array}$

Now on plug in the above value of y in this last equation,

  $\begin{array}{l}{\left(\frac{10+x}{2}\right)}^2=x(16+x)\\ \Rightarrow \frac{100+x^2+20x}{4}=x^2+16x\\ \Rightarrow 100+x^2+20x=4x^2+64x\\ \Rightarrow 4x^2+64x-100-x^2-20x=0\\ \Rightarrow 3x^2+44x-100=0\end{array}$

Factorize it as,

  $\begin{array}{l}3x^2+50x-6x-100=0\\ x(3x+50)-2(3x+50)=0\\ (3x+50)(x-2)=0\\ \Rightarrow 3x+50=0\\ \Rightarrow x=-\frac{50}{3}(\text{Rejected)}\end{array}$

And, $x-2=0\Rightarrow x=2$

Conclusion: Thus,by above calculation it is cleared that $ED=2$ .