## Math 9 Chapter 3 Lesson 4: Angle formed by a tangent ray and chord

## 1. Summary of theory

### 1.1. Concept

An angle formed by a tangent ray and a chord is an angle whose vertex lies on the circle, one side is a tangent ray and the other side contains the chord of the circle.

The inner arc is the blocked bow.

In the figure above we have: angle \(\widehat{BAx}\) (or \(\widehat{BAy}\)) is the angle formed by the tangent and the chord.

### 1.2. Theorem

The measure of the angle formed by the tangent ray and the chord is half the measure of the intercepted arc

Specifically in the image above, \(\widehat{BAx}=\frac {1}{2}\)sđ\(\stackrel\frown{AB}\) (here is the minor arc AB)

### 1.3. Consequences

In a circle, the angle formed by the tangent ray and the chord and the inscribed angle intercepting the same arc are equal.

## 2. Illustrated exercise

2.1. Basic exercises

**Question 1: **Given a circle \((O;R)\) and a point \(A\) on the circle, the tangent at \(A\) intersects the diameter \(BC\) of the circle at \(S\). Knowing \(\widehat{SAB}=30^0\), calculate \(AC\) in terms of \(R\).

**Solution guide**

We have \(\widehat{SAB}+\widehat{BAO}=90^0 \Rightarrow \widehat{BAO}=90^0-30^0=60^0\)

\(\bigtriangleup OBA\) at \(O\) has \(\widehat{BAO}=60^0\) so \(\bigtriangleup BAO\) is even. Derive \(BA=OB=R\)

Applying Pythagorean theorem to right triangle \(ABC\) we have \(AC=\sqrt{BC^2-AB^2}=\sqrt{(2R)^2-R^2}=\sqrt{3R ^2}=R\sqrt{3}\)

**Verse 2: **Let the circle \((O;R)\) and the point \(I\) lie outside the circle such that \(OI=2R\). The point \(C\) lies on the circle. Draw a tangent \(IA\) to the circle, let \(B\) be the intersection of \(OI\) and \((O)\) (\(B\) between \(O\) and \ (I\)). Calculate \(\widehat{ACB}\)

**Solution guide**

We have \(BI=OI-OB=2R-R=R\)

Right triangle \(AOI\) has \(B\) is the midpoint of \(OI\) so \(BA=BO=BI=R\) deduces \(\bigtriangleup OBA\) is regular (all sides are equal \(R\))

so \(\widehat{BOA}=60^0 \Rightarrow \widehat{ACB}=30^0\)

**Question 3: **Given triangle ABC, draw a circle with center O passing through A and tangent to BC at B. Draw the line BD parallel to AC. Let I be the intersection of CD with the circle. Proof: \(\widehat{IAB}=\widehat{ICA}=\widehat{IBC}\)

**Solution guide**

As a consequence of the theorem of the angle formed by the tangent and the chord, we have \(\widehat{IAB}=\widehat{IBC}=\widehat{IDB}\) (intercept \(\stackrel\frown{IB}\) )

Otherwise, \(\widehat{IDB}=\widehat{ICA}\) (do \(BD//AC\))

From (1) and (2) deduce \(\widehat{IAB}=\widehat{ICA}=\widehat{IBC}\) (dpcm)

### 2.2. Advanced exercises

**Question 1: **Given a circle \((O)\) and a point \(M\) outside the circle, from \(M\) draw a line \(MAB\) to the circle. \(C\) is the point on the other circle \(A\) and \(B\). Prove that: \(MC\) is a tangent to the circle \((O)\) if and only if \(MC^2=MA.MB\)

**Solution guide**

* Forward direction:* \(MC\) is tangent to the resulting circle \(\widehat{MCA}=\widehat{MBC}\)

Considering \(\bigtriangleup MAC\) and \(\bigtriangleup MCB\) have \(\widehat{M}\) in common and \(\widehat{MCA}=\widehat{MBC}\) so \(\bigtriangleup MAC \ sim \bigtriangleup MCB\) (gg)

deduce \(\frac{MA}{MC}=\frac{MC}{MB} \Rightarrow MC^2=MA.MB\)

* Reversal direction:* \(MC^2=MA.MB \Rightarrow \frac{MA}{MC}=\frac{MC}{MB}\)

Considering \(\bigtriangleup MAC\) and \(\bigtriangleup MCB\) have \(\widehat{M}\) in common and \(\frac{MA}{MC}=\frac{MC}{MB}\) so \(\bigtriangleup MAC \sim \bigtriangleup MCB\) (cgc)

deduce \(\widehat{MCA}=\widehat{MBC} \Rightarrow \widehat{MCA}=\frac{1}{2}\)sđ\(\stackrel\frown{AC}\)

Draw the diameter \(CD\) then \(\widehat{MCA}+\widehat{ACD}=\frac{1}{2}\)\(\widehat{MCD}=\widehat{MCA}+\widehat {ACD}=\frac{1}{2}\)sđ\(\stackrel\frown{AC}\)+\(\frac{1}{2}\)sđ\(\stackrel\frown{AD}\ )=\(90^0\)

From this it follows that \(MC\) is a tangent to the circle \((O)\)

**Verse 2: **Given two circles \((O)\) and \((O’)\) intersect at \(A\) and \(B\). The tangent at \(A\) to the circle \((O’)\) intersects \((O)\) at \(C\) and to the circle \((O)\) intersects \((O) ‘)\) at \(D\).

Prove \(AB^2=BD.BC\)

**Solution guide**

In the circle \((O)\) we have \(\widehat{ACB}=\widehat{BAD}\) (the inscribed angle and the angle formed by the tangent ray and the chord intercepting the arc \(BA\))

Similarly in the circle \((O’)\) we also have \(\widehat{BDA}=\widehat{BAC}\)

Consider \(\bigtriangleup CAB\) and \(\bigtriangleup ADB\) where \(\widehat{ACB}=\widehat{BAD}\) and \(\widehat{BDA}=\widehat{BAC}\)

so \(\bigtriangleup CAB\sim \bigtriangleup ADB\) infers \(\frac{CB}{AB}=\frac{AB}{DB}\Rightarrow AB^2=BD.BC\)

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Two circles (O) and (O’) intersect at A and B. Through A draw a CAD line with two circles (C ∈ (O) ,D ∈ (O’))

a. Prove that when the sand line rotates around point A, then have constant measure

b. From C and D draw two tangents to the circle. Prove that these two tangents join at an angle of constant measure when the CAD sandline rotates around point A.

**Verse 2: **From a fixed point M outside the circle (O), draw a tangent MT and a tangent MAB to that circle.

a. Prove that there is always MT2= MA.MB and this product does not depend on the position of the sandline MAB

b. Let MT = 20cm ,MB = 50cm, calculate the radius of the circle

**Question 3:** Sitting on the top of a mountain 1 km high, what is the maximum distance that a point T can be seen on the ground? Knowing that the radius of the Earth is close to 6400 km

**Question 4: **Let ABC be a triangle inscribed in circle (O). Draw ray Bx such that ray BC lies between two rays Bx, BA and Prove that Bx is a tangent to (O).

### 3.2. Multiple choice exercises

**Question 1: **Let ABC be a triangle inscribed in circle (O). The tangent at A intersects BC at I. Know AB=20cm, AC=28cm, BC=24cm. Then IA equals how many cm?

A. 32cm

B. 20cm

C. 28cm

D. 35cm

**Verse 2: **Which of the following statements is true:

A. The angle formed by the tangent ray and the chord is always less than 900.

B. The angle formed by the tangent ray and the chord and the inscribed angle are always equal.

C. The measure of the angle formed by the tangent ray and the chord is half the measure of the angle at the center of the same arc.

D. The angle formed by two chords of a circle is always less than 900.

**Question 3: **Which of these following statements is wrong:

A. The measure of the angle formed by the tangent ray and the chord is half the measure of the intercepted arc

B. In a circle, the measure of the angle formed by the tangent ray and the chord and the inscribed angle intercepting the same arc are equal.

C. If angle BAx has half the measure of arc BA outside that angle, then Ax is a tangent to the circle containing arc AB.

D. With MAB being the tangent and T being a point on the circle, from the formula MT2=MA.MB we infer that MT is the tangent to that circle.

**Question 4: **Given two circles \((O)\) and \((O’)\) intersect at \(A\) and \(B\). The tangent at \(A\) to the circle \((O’)\) intersects \((O)\) at \(C\) and to the circle \((O)\) intersects \((O) ‘)\) at \(D\). Know that BC=16cm, BD=12cm. The length BA is:

A. \(8\sqrt{3} cm\)

B. \(10 cm\)

C. \(8\sqrt{2} cm\)

D. \(8 cm\)

**Question 5: **Given a circle (O;R), point A lies outside the circle and OA=2R. From A draw two tangents AB and AC to the circle (O). Let H be the intersection of BC and OA. Then which of the following is false:

A. \(\bigtriangleup ABC\) are all

B. \(AO\perp BC\)

C. \(HO=\frac{R}{2}\)

D. \(BC=R\sqrt{2}\)

## 4. Conclusion

Through this lesson, you will learn some key topics as follows:

- Identify the angle formed by the tangent ray and the chord.
- State and prove the theorem about the measure of the angle formed by the tangent ray and the chord in 3 cases.
- Students can divide the cases to conduct the proof.

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