Q19. - Geometry

Question

Given: Two tangents circles; $\overline{EF}$ is a common external tangent;

                $\overline{GH}$ is the common internal tangent.

$\left(a\right).$ Discover and prove something interesting about point $G.$

$\left(b\right).$ Discover and prove something interesting about $\angle EHF$.

Step-by-Step Solution

Answer

a. Point $G$ is the midpoint of $EF$.

b. Value of, $m\angle EHF={90}^0$.

Part a. Step 1. Given information.

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 $G$.

Step 2. Concept used.

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)

Step 3. Concluding from equation (i) and (ii).

We get,

$EG=FG$

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

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

Part b. Step 1. Given information.

Two tangents circles; $EF$ is a common external tangent; 

$GH$ is the common internal tangent.

Step 2. Draw the construction.

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 $EHF$.

Step 3. Use 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.

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

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