Chapter 3.7, Problem 23PSC

To determine

To prove : AOBP is equilateral.

Explanation of Solution

Given information : The following information has been given

O is the center of first circle

P is the center of second circle

  $\overrightarrow{AB}$ bisects $\angle OAP$ and $\angle OBP$

Formula used : Any chord drawn makes an isosceles triangle with the center of the circle

Proof : If O is the center of first circle , then we can say that

  $\left|\overrightarrow{OA}\right|\cong \left|\overrightarrow{OB}\right|$

Also, if P is the center of second circle, then we can say that

  $\left|\overrightarrow{PA}\right|\cong \left|\overrightarrow{PB}\right|$

Now, if $\overrightarrow{AB}$ bisects $\angle OAP$ and $\angle OBP$ , then we can say that

  $\angle OBA\cong \angle PBA$

  $\angle OAB\cong \angle PAB$

Now, in triangles ABO and ABP, we can say that

  $\angle OBA\cong \angle PBA$

  $\angle OAB\cong \angle PAB$

AB is the common chord

Thus, by ASA property, we can say that $\Delta ABO\cong \Delta ABP$

Now, as $\Delta ABO\cong \Delta ABP$ , we know that

  $\left|\overrightarrow{OA}\right|\cong \left|\overrightarrow{PA}\right|$

Now, combining the previous equation with

  $\left|\overrightarrow{OA}\right|\cong \left|\overrightarrow{OB}\right|$ and $\left|\overrightarrow{PA}\right|\cong \left|\overrightarrow{PB}\right|$ , we get

  $\left|\overrightarrow{PA}\right|\cong \left|\overrightarrow{PB}\right|\cong \left|\overrightarrow{OA}\right|\cong \left|\overrightarrow{OB}\right|$

Hence, proved that AOBP is equilateral