# Textbooks for High School Students Studying the Mathematics

# Textbooks for High School Students Studying the Mathematics

**Exercise: Circles IV
**1. Find the values of the unknown letters.

**Theorem 14.**Two tangents drawn to a circle from the same point outside the circle are equal

in length.

**Proof:**

Consider a circle, with centre O. Choose a point P outside the circle. Draw two tangents to the

circle from point P, that meet the circle at A and B. Draw lines OA , OB and OP.

**The aim is to prove that AP = BP.**

In △OAP and △OBP,

1. OA = OB (radii)

2. OAP =OPB = 90°(OA ⊥ AP and OB ⊥ BP)

3. OP is common to both triangles.

△OAP ≡ △OBP (right angle, hypotenuse, side)

AP = BP

**Exercise: Circles V
**1. Find the value of the unknown lengths.

**Theorem 15.**The angle between a tangent and a chord, drawn at the point of contact of the

chord, is equal to the angle which the chord subtends in the alternate segment.

**Proof:**

Consider a circle, with centre O. Draw a chord AB and a tangent SR to the circle at point B.

Chord AB subtends angles at points P and Q on the minor and major arcs, respectively.

Draw a diameter BT and join A to T .

**The aim is to prove that
and .**

First prove that as this result is needed to prove that

**Exercise: Circles VI
**1. Find the values of the unknown letters.

**Theorem 16.**(Converse of 15) If the angle formed between a line, that is drawn through the

end point of a chord, and the chord, is equal to the angle subtended by the chord in the alternate

segment, then the line is a tangent to the circle.

**Proof:**

Consider a circle, with centre O and chord AB. Let line SR pass through point B. Chord AB

subtends an angle at point Q such that

**The aim is to prove that SBR is a tangent to the circle.**

By contradiction. Assume that SBR is not a tangent to the circle and draw XBY
such that

XBY is a tangent to the circle.

If is zero , then both XBY and SBR coincide and SBR is a tangent to the
circle.

**Exercise: Applying Theorem 9
**1. Show that Theorem 9 also applies to the following two cases:

**Worked Example 190: Circle Geometry I**

Question:

Question:

BD is a tangent to the circle

with centre O.

BO ⊥ AD.

Prove that:

1. CFOE is a cyclic quadrilateral

2. FB = BC

3. △COE///△CBF

4. CD^{2} = ED.AD

**Answer
1. Step 1 : To show a quadrilateral is cyclic, we need a pair of opposite
angles to be supplementary, so lets look for that.
** ( subtended by diameter AE)

CFOE is a cyclic quadrilateral (opposite 's supplementary)

**2. Step 1 : Since these two sides are part of a triangle, we are proving that
triangle to be isosceles. The easiest way is to show the angles opposite
to those sides to be equal.
**Let

( between tangent BD and chord CE)

(exterior to cyclic quadrilateral CFOE)

BF = BC (sides opposite equal 's in isosceles △BFC)

**3. Step 1 : To show these two triangles similar, we will need 3 equal angles.
We already have 3 of the 6 needed angles from the previous question .
We need only find the missing 3 angles.
**

**4. Step 1 : This relation reminds us of a proportionality relation between
similar triangles. So investigate which triangles contain these sides and
prove them similar. In this case 3 equal angles works well. Start with
one triangle.
**In △EDC

('s on a straight line AD)

(complementary 's)

**Step 2 : Now look at the angles in the other triangle.
**In △ADC

( sum of 's and )

(sum of 's in △CAE)

**Step 3 : The third equal angle is an angle both triangles have in common.
**Lastly, since they are the same

**Step 4 : Now we know that the triangles are similar and can use the
**proportionality relation accordingly.

**5. Step 1 : This looks like another proportionality
relation with a little
twist, since not all sides are contained in 2 triangles. There is a quick
observation we can make about the odd side out, OE.
**OE = CD (△OEC is isosceles)

**Step 2 : With this observation we can limit ourselves to proving triangles
BOC and ODC similar. Start in one of the triangles.
**In △BCO

(radius OC on tangent BD)

(sum of 's in △BFC)

**Step 3 : Then we move on to the other one.
**In △OCD

(radius OC on tangent BD)

(sum of 's in △OCE)

**Step 4 : Again we have a common element .
**Lastly, OC is a common side to both △'s.

**Step 5 : Then, once we’ve shown similarity, we use the proportionality
relation , as well as our first observation, appropriately.
**

**Worked Example 191: Circle Geometry II
Question:
**

FD is drawn parallel to the

tangent CB

Prove that:

1. FADE is cyclic

2. △AFE///△CBD

**Answer
1. Step 1 : In this case, the best way to show FADE is a cyclic quadrilateral
is to look for equal angles, subtended by the same chord.
**Let
BCD = x

CAH = x ( between tangent BC and chord CE)

FDC = x (alternate , FD || CB)

FADE is a cyclic quadrilateral (chord FE subtends equal 's)

**2. Step 1 : To show these 2 triangles similar we will need 3 equal angles.
We can use the result from the previous question.
**Let
FEA = y

FDA = y ('s subtended by same chord AF in cyclic quadrilateral FADE)

CBD = y (corresponding 's, FD || CB)

FEA =CBD

**Step 2 : We have already proved 1 pair of angles equal in the previous
question.
**
BCD
=
FAE (above)

**Step 3 : Proving the last set of angles equal is simply a matter of adding
up the angles in the triangles. Then we have proved similarity.
**
AFE
= 180°-x - y ('s in △AFE)

CBD = 180° x - y ('s in △CBD)

△AFE///△CBD (3's equal)

**3. Step 1 : This equation looks like it has to do with proportionality relation
of similar triangles. We already showed triangles AFE and CBD similar
in the previous question. So lets start there.
**

**Step 2 : Now we need to look for a hint about side FA. Looking at
triangle CAH we see that there is a line FG intersecting it parallel to
base CH. This gives us another proportionality relation .
**(FG
|| CH splits up lines AH and AC proportionally)

**Step 3 : We have 2 expressions for the side FA.
**

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