Chapter 9.2, Problem 19WE

a.

To determine

To discover: prove something interesting about point $G$ .

Answer to Problem 19WE

Point $G$ is the midpoint of $EF$

Explanation of Solution

Given:

Two tangents circles; $EF$ is a common external tangent; $GH$ is the common internal tangent.

  

For the larger circle, EG and GH are two tangents drawn from a common point G.

By the property of tangents, if two tangents of the circle are drawn from a common point, then their length are equal.

Thus, $EG=GH$ …(i)

Now, for the smaller circle, GH and FG are two tangents drawn from a common point G.

By the property of tangents, if two tangents of the circle are drawn from a common point, then their length are equal.

Thus, $FG=GH$ …(ii)

Concluding from equation (i) and (ii), $EG=FG$

Hence, G is the midpoint of the common external tangent EF of both the circle.

Conclusion:

Therefore, $G$ is the midpoint of external tangent $EF$ .

b.

To determine

To discover: prove something interesting about $\angle EHF$ .

Answer to Problem 19WE

  $m\angle EHF={90}^0$

Explanation of Solution

Given:

Two tangent circles; $EF$ is a common eternal tangent; $GH$ is the common internal tangent.

  

Since $EG=GH=GF$

A circle can be constructed with G as the center and $EG=GH=GF$ as radius.

  

In the circle with center G, EF is the diameter and angle $\angle EHF$ is made in the semicircular arc EHF.

By the property of circle, any angle made in a semicircular arc is always a right angle, it can be easily said that $\angle EHF$ is a right angle.

Therefore, $\Delta EHF$ is a right angled triangle and $\angle EHF={90}^0.$

Conclusion:

Therefore, $m\angle EHF={90}^0$ .