## Tangent segment and secant of a circle, Arc angle subtending concept and External angle property of a chord

The square of a tangent segment from an external point will be equal to the product of the two segments of a secant of a circle through the point. This important relation between tangent segments and secant of a circle is explored.

In this 9th concept tutorial session on Geometry, the important relation between the secant segments and the tangent segment from the external point is explored in details with proofs.

To prove the **secant segment relation**, two more basic and rich geometry concepts need to be used,

**Arc angle subtending concept**in a circle, and,**External angle equality property of a chord**in a circle.

These two concepts will be stated and proved first to explain the proof of secant segment relation clearly.

The three concepts are covered in separately earlier tutorials,

* Basic and rich Geometry concepts 4, Detailed proof of Arc angle subtending concept,* and,

**Basic and rich Geometry concepts part 3, Circles.**

But In this session *we'll explain the three concepts together keeping the explanation brief.*

### Arc angle subtending concept in a Circle and its short proof

We'll use the following graphic for stating and explaining the concept.

The important arc angle subtending concept in a circle states,

An arc segment in a circle subtends the same angle at every point on the alternate arc segment of the circle, and at the centre of the circle, the arc subtends an angle double the angle it subtends on the alternate segment of the periphery.

Let us see how this property comes to be true.

The small segment of the arc AB subtends an angle $\angle \theta$ at P on the alternate segment of the periphery.

Being the radii, $AO=PO$ and in isosceles $\triangle AOP$, the two base angles,

$\angle PAO=\angle APO=x$, say.

Similarly, in isosceles $\triangle BOP$, the two base angles,

$\angle PBO=\angle BPO=y$, say.

So, $\angle APB = x+y=\theta$, say.

Now in $\triangle AOB$,

$\angle AOB=180^0-(\angle OAB+\angle OBA)$.

In larger $\triangle PAB$,

$(\angle OAB+\angle OBA)=180^0-2(x+y)$.

So,

$\angle AOB=180^0-[180^0-2(x+y)]=2(x+y)=2\theta$.

The same is true for $\angle AQB$ subtended by the arc AB at a second point Q (and any other point) on the alternate arc segment of the periphery.

At all points on the alternate arc segment of the periphery, the arc AB holds the same angle $\theta$ that is half the angle it holds at the centre.

### External angle equality property of a chord and its short proof

The following graphic will be used for stating and proving the concept.

SA is the tangent to the circle at point A where it is perpendicular to the radius OA.

The $\angle SAB$ held by the chord BA at tangent point A with the tangent segment SA is called the external angle of the chord.

The **external angle property of a chord** states,

The external angle of a chord will be equal to the angle held by the chord at any point on the alternate segment of the periphery. To be specific, $\angle SAB=\angle APB=\theta$.

Let us understand how this result comes true.

Let, $\angle SAB=\theta$.

As radius OA is perpendicular to SA at tangent point A,

$\angle SAB+\angle BAO=90^0$,

Or, $\angle BAO=\angle ABO=90^0-\theta$, as radii $OA=OB$ in isosceles $\triangle OAB$.

So in $\triangle OAB$,

$\angle AOB=180^0-2[90^0-\theta]=2\theta$.

This being the angle held by the chord AB at the centre, by arc angle subtending property, it is double the angle held by the same chord AB at the point P or Q on the alternate segment of the corresponding arc AB on the periphery.

So, $\angle SAB=\angle APB=\angle AQB=\theta$.

With the two concepts understood, we are now ready to explain the main **secant segment target concept.**

### Segment relation of a secant with the tangent segment of a circle

In short we call this relation as secant segment relation. The following graphic will be used for stating and proving the important property.

The **secant segment property** states,

The square of a tangent segment from an external point will be equal to the product of the two segments of a secant of a circle through the point.

In the graphic, AP is the tangent segment from external point A and ABC is the secant of the circle through point A. The circle segments the secant into AB and BC. By the secant segment relation,

$AP^2=AB.AC$.

Let us see how this property comes true.

Observe that by external angle property of a chord, $\angle APB=\angle ACP$ in two triangles $\triangle ABP$ and $\triangle ACP$.

With $\angle A$ common, the third pair of angles are equal,

$\angle ABP=\angle APC$.

With all three pairs of angles equal, **by A-A-A property**, the two triangles are similar.

In a pair of similar triangles, **ratio of corresponding pairs of sides are equal.** So by this property,

$\displaystyle\frac{AP}{AC}=\frac{AB}{AP}$,

Or, $AP^2=AB.AC$.

In the first ratio of LHS, AP in numerator and AC in denominator are opposite to equal angles $\angle ABP$ in $\triangle ABP$ and $\angle APC$ in $\triangle APC$ respectively.

Similarly, in the second RHS ratio, AB in numerator and AP in denominator are opposite to equal angles $\angle APB$ in $\triangle ABP$ and $\angle ACP$ in $\triangle APC$ respectively.

These form the **corresponding pairs of sides.**

Observe how the proving of the third concept needed the prior understanding and use of the other two concepts.

### Guided help on Geometry in Suresolv

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