Chapter 10.6, Problem 22PSC

a.

To determine

Give the explanation for the given proof.

Explanation of Solution

Given:

The given diagram is

  

$\overline{AC}$ and $\overline{BD}$ intersecting at $E$ . The given proof is $mArc\left(AB\right)+mArc\left(CD\right)=2\left(m\angle CED\right)$ .

Calculation:

Draw a segment $\overline{AD}$ and $\overline{BC}$ .

  

The inscribed angles which intercepted by the same arc are congruent.

Use $Arc\left(AB\right)$ .

  $\angle ADE\cong \angle BCE=x\left(\text{let}\right)$

Use $Arc\left(CD\right)$ .

  $\angle DAE\cong \angle CBE=y\left(\text{let}\right)$

Inscribed angle $=\frac{1}{2}$ intercepted arcs

  $\begin{array}{l}m\angle ADE=\frac{1}{2}mArc\left(AB\right)\\ \Rightarrow x=\frac{1}{2}mArc\left(AB\right)\\ \Rightarrow 2x=mArc\left(AB\right)\end{array}$

Similarly $mArc\left(CD\right)=2y$

In $\Delta CBE$ .

  $\angle CEB={180}^{°}-\left(x+y\right)$

Now,

  $\begin{array}{l}m\angle CED={180}^{°}-\left[{180}^{°}-\left(x+y\right)\right]\\ ={180}^{°}-{180}^{°}+\left(x+y\right)\\ =x+y\end{array}$

From LHS

  $\begin{array}{l}mArc\left(AB\right)+mArc\left(CD\right)\\ =2x+2y\\ =2\left(x+y\right)\\ =2\left(m\angle CED\right)=\text{RHS}\end{array}$

Hence $mArc\left(AB\right)+mArc\left(CD\right)=2\left(m\angle CED\right)$ proved.

b.

To determine

Give the explanation for the given proof.

Explanation of Solution

Given:

The given diagram is

  

$\overline{AC}$ and $\overline{BD}$ intersecting at $E$ . The given proof is $AE\cdot EC=BE\cdot ED$ .

Calculation:

Draw a segment $\overline{AD}$ and $\overline{BC}$ .

  

The inscribed angles which intercepted by the same arc are congruent.

Use $Arc\left(AB\right)$ .

  $\angle ADE\cong \angle BCE=x\left(\text{let}\right)$

Use $Arc\left(CD\right)$ .

  $\angle DAE\cong \angle CBE=y\left(\text{let}\right)$

In $\Delta CBE$ .

  $\angle CEB={180}^{°}-\left(x+y\right)$

In $\Delta ADE$ .

  $\angle AED={180}^{°}-\left(x+y\right)$

  $\angle CEB=\angle AED$

From AA similarity.

  $\Delta CBE\simeq \Delta ADE$

Now,

  $\begin{array}{l}\frac{AE}{ED}=\frac{BE}{EC}\\ \Rightarrow AE\cdot EC=BE\cdot ED\end{array}$

Hence $AE\cdot EC=BE\cdot ED$ proved.