What Is the Ratio of Triangles to Circles

Triangles and Circles

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

Statements Of Some Theorems On The Circle.

Theorem A
A straight line fatigued from the centre of a circle to bisect a chord which is non a diameter, is at right angles to the chord.

A chord is a line segment joining 2 points on a circle.

Theorem B
There is only one circle which passes through three given points which are non in a straight line.

Theorem C
Equal chords of a circle are equidistant from the heart and visa versa.

A tangent is a line intersecting the circle at simply 1 bespeak.

Theorem D
The tangent to a circle and the radius through the point of contact are perpendicular to each other.

Theorem E
The angle which an arc of a circle subtends at the centre is double that which it subtends at whatsoever bespeak on the remaining part of the circumference.

Theorem F
Angles in the same segment of a circumvolve are equal.

Theorem Thou
The bending in a semicircle is a right angle.

Theorem H
The opposite angles of whatever quadrilateral inscribed in a circumvolve are supplementary.

Theorem I
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 alternating segment.

Instance - Supplementary

Statements Of Some Theorems On Proportions And Similar Triangles.

Theorem J
If a directly line is drawn parallel to i side of a triangle, the other 2 sides are divided proportionally.

Theorem Yard
If two triangles are equiangular their corresponding sides are proportional.

Theorem 50
If two triangles have one equal angle and the sides about these equal angles are proportional, then the triangles are similar.

Theorem G
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 North
The internal bisector of an angle of a triangle divides the reverse side in the ratio of the sides containing the bending.

Example - Example 1

Trouble

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

Workings

Proof

Since AP is parallel to BC

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

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

Hence triangle BHQ is equiangular with triangle APH

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

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

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

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

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

Combining the iii above equations:-

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

Solution

Conclusion

Information technology can be seen that $\inline BQ$ = $\inline CR$ $\inline Q.E.D.$

Pythagoras Theorem

Theorem m provides a user-friendly method of proving Pythagoras theorem. Consider the right angled triangle $\inline ABC$ .

Pytagoras theorem
Given the right triangle $\inline \triangle ABC$ prove that $\inline BC^2=AB^2+AC^2$ .

Structure.
Draw $\inline AD$ such that the bending $\inline ADB$ is a correct angle.

Proof
The triangles $\inline ABC$ : $\inline ABD$ and $\inline ADC$ are equiangular and similar. From

$\triangle\;ABC\; \sim\triangle\;ADC\;\;\;\;\;\frac{BC}{AC} = \frac{AC}{CD}$

therefore

$AC^2 = BC\times CD$

From

$\triangle\;ABC\;\sim\triangle\;ABD\;\;\;\;\;\frac{BC}{AB} = \frac{AB}{DB}$

therefore

$AB^2 = CB\times DB$

Add equations (22) and (24)

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

therefore

$\mathbf{AB^2 + AC^2 = BC^2}$

Note
$\inline \sim$ ways that the two triangles are similar .

Two Theorems On Similar Rectilinear Figures.

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

Theorem i

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

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

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

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

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

If Polygons are like they tin be divided upwards into the aforementioned number of like triangles and it follows that the ratio of the areas of similar polygons is equal to the ratio of the squares on corresponding sides.

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 :

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

So the polygons ABCD.....P, A'B'C'D'.....P' are like and similarly situated.

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

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

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

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 ii 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 center 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 center of similitude.

Instance - Similarities

Problem

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

Workings

Solution

And so $\inline ABCD$ is the required square for regarding $\inline P$ as the eye of Similitude $\inline ABCD$ is similar to $\inline QHKR$ and is therefore a square.

Concluding Modified: 29 May 11 @ ten:00 Page Rendered: 2022-03-xiv 15:53:07

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Source: https://www.codecogs.com/library/maths/geometry/pure/triangles-and-circles.php

What Is the Ratio of Triangles to Circles

Triangles and Circles

Statements Of Some Theorems On The Circle.

Instance - Supplementary

Statements Of Some Theorems On Proportions And Similar Triangles.

Example - Example 1

Pythagoras Theorem

Two Theorems On Similar Rectilinear Figures.

Instance - Similarities

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