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Theorems on Circles and Triangles including a proof of the Pythagoras Theorem

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Statements of some theorems on the Circle.

A straight line drawn from the centre of a Circle to bisect a chord which is not a diameter, is at right angles to the Chord. (theorem a)

There is only one circle which passes through three given points which are not in a straight line. (theorem b)

Equal chords of a circle are equidistant from the centre and visa versa. (theorem c)

The tangent to a circle and the radius through the point of contact are perpendicular to each other. (theorem d)

The Angle which an arc of a circle subtends at the centre is double that which it subtends at any point on the remaining part of the circumference. (theorem e)

Angles in the same segment of a circle are equal (theorem f)

The angle in a semicircle is a right angle. (theorem g)

The opposite angles of any quadrilateral inscribed in a circle are Supplementary. (theorem h)

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If a straight line touches a circle and from the point of contact a chord is drawn, the angles which this tangent makes with the chord are equal to the angles in the alternate segment. (theorem i)

Example 1

To Prove A line DE parallel to the base BC of the triangle ABC cuts AB, AC at D and E respectively. The circle which passes through D and touches AC at E meets AB at F. Prove that F,E,C,B, lie on a circle.

Construction Draw the line FE

Proof

$Angle\;CED = Angle\;EFD \;\;\;\;(Theorem\;i)$

(1)

Since DE is parallel to BC the angles CED and BCA are Supplementary

Hence the angles BCA and EFD are supplementary

Conclusion from Theorem h above the points F,E,C,B. lie on a circle. Q.E.D.

Example 2

In a triangle ABC, the side AB is greater than the side AC and D is a point on AB such that AD = AC. The internal bisectors of the angles B and C meet at I. Show that the four points B,D,I,C. lie on a circle.

Proof Since AD = AC the triangle ADC is Isosceles.

$\therefore\;\;\;\;\;\;\;\;Angle\; ADC = Angle\;DCA$

(2)

The sum of the internal angles of a triangle is 180 degrees

$\therefore\;\;\;\;\;\;\;\;2\times angle\; ADC + angle \;CAD = 180^0$

(3)

$or\;\;\;\;\;\;\;\;angle\;ADC = 90^0 - \frac{1}{2}\;angle\;CAD$

(4)

So the angle CDB is the supplement of angle ADC

$\therefore\;\;\;\;\;\;\;\;angle\;CDB = 180^0\;-(90^0 - \frac{1}{2}A) = 90^0 + \frac{1}{2}\;angle \;CAD$

(5)

$since \;angle\;IBC = \frac{1}{2}\;angle\; DBC\;\;\;and\;\;\;angle\; BCI = \frac{1}{2}\;angle\;BCI$

(6)

From the triangle IPC it can be seen that :-

$angle\;CIB = 180^0 - \frac{1}{2}\;angle\;ABC - \frac{1}{2}\;ACB$

(7)

$= 180^0 - \frac{1}{2}(180^0 - angle\;CAD) = 90^0 + \frac{1}{2}\;angle\; CAD$

(8)

$\therefore\;\;\;\;\;\;angle\;CDB = angle\;CIB$

(9)

As these two angles are equal it can be seen that they satisfy the converse of Theorem (f).

Conclusion The points B,D,I,C. lie on a circle Q.E.D.

Statements of some Theorems on Proportions and Similar Triangles.

If a straight line is drawn parallel to one side of a triangle, the other two sides are divided proportionally. (theorem j)

If two triangles are equiangular their corresponding sides are proportional. (theorem k)

If two triangles have one equal angle and the sides about these equal angles are proportional, then the triangles are similar. (theorem l)

If a triangle is drawn from the right angle of a right angled triangle to the hypotenuse, then the triangles on each side of of the perpendicular are similar to the whole triangle and to one another. (theorem m)

The internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle. (theorem n)

Example 3

Any line parallel to the base BC of a triangle ABC cuts AB and AC in H and K respectively. P is any point on on a line through A parallel to BC. If PH and PK produced cut BC at Q and R respectively, Prove that BQ = CR.

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Proof

Since AP is parallel to BC

$angle\;HAP = angle\;HBQ\;\;\;\;\;\;\;\;(alternate\; angles)$

(10)

$Also \;angle\;BHQ = angle\;PHA$

(11)

Hence triangle BHQ is equiangular with triangle APH

$\frac{BQ}{AP} = \frac{BH}{AH}\;\;\;\;\;\;\;\;(similar\;triangles)$

(12)

Likewise the triangles APK and KCR are equiangular and hence:-

$\frac{BH}{AH} = \frac{CK}{AK}$

(13)

Since HK is parallel to BC, theorem (k) applies and we can write:-

$\frac{BH}{AH} = \frac{CK}{AK}$

(14)

Combining the three above equations:-

$\frac{BQ}{AP} = \frac{CR}{AR}$

(15)

From which it can be seen that BQ = CR Q.E.D.

Example 4

ABC is a triangle right angled at A. AN is perpendicular to BC. BK bisects the angle ABC and meets AC at K and AN at L.

To Prove AL:LN = CK:KA

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Proof

Applying theorem (n) to triangle ABN

$\frac{AL}{LN} = \frac{BA}{BN}$

(16)

Similarly for triangle ABC:-

$\frac{CK}{KA} = \frac{BC}{BA}$

(17)

But theorem (m) shows that the triangles ABN and ABC are similar

$\frac{BA}{BN} = \frac{BC}{BA}$

(18)

Combining the three last equations:-

$\frac{AL}{LN} = \frac{CK}{KA}\;\;\;\;\;\;\;\;Q.E.D.$

(19)

Pythagoras's Theorem

Theorem m provides a convenient method of proving Pythagoras's theorem. Consider the right angled triangle ABC.

$\mathbf{To\;Prove\;that\;BC^2 = AC^2 + AB^2}$

(20)

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Construction. Draw AD such that the angle ADB is a right angle.

Proof The triangles ABC : ABD and ADC are equiangular and similar.

$From\;triangles\;ABC\; and\;ADC\;\;\;\;\;\frac{BC}{AC} = \frac{AC}{CD}$

(21)

$\therefore\;\;\;\;\;\;AC^2 = BC\times CD$

(22)

$From\;triangles\;ABC\;and\;ABD\;\;\;\;\;\frac{BC}{AB} = \frac{AB}{DB}$

(23)

$\therefore\;\;\;\;\;\;AB^2 = CB\times DB$

(24)

Add equations (22) and (24)

$AC^2 + AB^2 = BC\times CD + BC\times DB = BC\left(CD + DB \right)$

(25)

$\mathbf{\therefore\;\;\;\;\;AB^2 + AC^2 = BC^2}\;\;\;\;\;\;\;Q.E.D.$

(26)

Two Theorems on Similar Rectilinear Figures.

Polygons which are equiangular and have their corresponding sides proportional are said to be similar. If also their corresponding sides are parallel, they are said to be similarly situated (or homothetic)

Theorem 1

The ratio of the areas of similar triangles (or polygons) is equal to the ratio of the squares on corresponding sides.

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ABC and PQR are similar triangles and AD and PS are their heights. Since the angle ABD equals angle PQS and the angle BDA equals angle QSP, the triangles ABD and PQS are equiangular and Hence:-

$\frac{AD}{PS} = \frac{AB}{PQ} = \frac{BC}{QR}$

(27)

The last equality follows from the fact that the triangles ABC and PQR are similar

$\therefore \;\;\;\;\;\frac{\Delta ABC}{\Delta PQR} = \frac{\frac{1}{2}AD\times BC}{\frac{1}{2}PS\times QR} = \frac{BC^2}{QR^2}$

(28)

If Polygons are similar they can be divided up into the same number of similar triangles and it follows that the ratio of the areas of similar polygons is equal to the ratio of the squares on corresponding sides.

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Theorem 2

If O is any fixed point and ABCD.....P is any polygon and if points A'B'C'.....P' are taken on OA,OB,OC,....OP ( or these lines produced in either direction) such that :-

$\frac{OA'}{OA} = \frac{OB'}{OB}\;=....=\frac{OP'}{OP} = \lambda$

(29)

Then the polygons ABCD.....P, A'B'C'D'.....P' are similar and similarly situated.

$Since\;\;\;\;\;\;\frac{OA'}{OA} = \frac{OB'}{OB}$

(30)

AB is parallel to A'B' and the triangles OAB and OA'B' are similar and hence:-

$\frac{A'B'}{AB} = \frac{OA'}{OA} = \lambda$

(31)

Corresponding sides of the two polygons are therefore proportional and parallel and the two polygons are similar and similarly situated.

In the above diagram "O" is said to be the centre of similitude of the two polygons. If the corresponding points of the two polygons lie on the same side of O the Polygons are said to be directly homothetic with respect to O and O is said to be the external centre of similitude. If the corresponding points lie on opposite sides of O then the Polygons are said to be inversely homothetic with respect to O and O is called the internal centre of similitude.

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Example

PQR is an acute angle triangle. Show how to construct a square with two vertices on QR and one vertex on PQ and one on PR.

Draw a square QHKR on the other side of QR to the triangle. Join PH and PK and let these lines meet QR at the points B and C respectively. Draw BA and CD perpendicular to QR to meet PQ and PR at the points A and D respectively. Then ABCD is the required square for regarding P as the centre of Similitude ABCD is similar to QHKR and is therefore a square.

Triangles and Circles

Statements of some theorems on the Circle.

Example 1

Example 2

Statements of some Theorems on Proportions and Similar Triangles.

Example 3

Example 4

Pythagoras's Theorem

Two Theorems on Similar Rectilinear Figures.