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A DIFFERENT DEMONSTRATION OF THE PYTHAGOREAN THEOREM USING THE POWER OF THE POINT WITH RESPECT TO A CIRCLE

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A DIFFERENT DEMONSTRATION OF THE PYTHAGOREAN THEOREM USING THE POWER OF THE POINT WITH RESPECT TO A CIRCLE



In the present paper I will give / offer you not only a different demonstration of the Pythagorean Theorem but also some demonstrations of other important theorems in geometry (The Height Theorem, The Catheters Theorem and The Generalized Pythagorean Theorem), using “The Power of the Point with Respect to a Circle”.

I would also like to add that these demonstrations are completely different from those already known.

THE POWER OF A POINT WITH RESPECT TO A CIRCLE

In this section I will point out / bring to mind what the power of a point with respect to a circle is.

STATEMENT 1. Any point P inside a circle and chord AB which contains this point, the product

PA ∙ PB is constant.

Demonstration. Given a point P inside a circle of centre O and ray / radius R and the variable chord AB, so that P (AB). We build / draw a chord CD through point P, see figure above.

Angles and have equal measures with half measure of the arc , we obtain

(1)

Angles APC and BPD being vertically opposed angles, are congruent.

(2)

From (1) and (2), we obtain: ,

we obtain / get the proportion , which is equivalent to the

statement .

The constant value of the product multiplied by (-1) is called the power of an internal point P with respect to the circle.

STATEMENT 2. Any point P outside a circle and the secant which contains this point and intersects the circle in points A and B (A(PB)), the product

is constant.

Demonstration. Given a point P outside the circle of centre O and ray / radius R and the secant PA which intersect the circle in A and B. We build a secant PC through point P, which intersects the circle in C and D, see figure above.

From m() + m() =

and

m() + m() = ,

we deduce that

m() = m(), resulting in

We have

and

,

From the similarity of the two triangles we obtain the proposition:

i.e. .

The constant value of the product is called the power of an external point P with respect to the circle.

Obs. If / As , then the straight line P is tangent to the circle and

becomes

2. THE PYTHAGOREAN THEOREM

“If in a triangle ABC, right angled in A, then ”.

Demonstration. We consider the triangle ABC a right-angled one, inscribed in the circle with the center O and radius R. Through O, the middle of the hypotenuse BC, we draw the parallel line OD for AB, D(AC) which intersects the circle in the points E and F; D(OE). As [OD] is a midline in the triangle ABC, we deduce that . Considering the power of the interior point D towards the circle, we have DA. , resulting that (1)

Taking into account that and that , the equality (1) becomes , and we obtain .

* This demonstration is different from the ones we know and it appeared for the first time in the magazine “Scoala Valceana” – The Mathematics notebook of the students in Valcea – edited by The School Inspectorate of Valcea County, The Society for Mathematical Sciences – Rm. Valcea subsidiary, Teachers Assembly house in Valcea, year 19…

THE RECIPROCAL PYTHAGOREAN THEOREM



If we have in a triangle ABC, then the angle A is a right one.

Demonstration. We consider the triangle ABC in which , inscribed in the circle with the center O and radius R. Through M, the middle of the side BC, we draw the parallel line MD for AB, which intersects the circle in E and F; . As [MD] is midline in the triangle ABC, we deduce that . Considering the power of the interior points M and D towards the circle, we have (1) and respectively (2).

The equality no. (2) becomes , resulting in (3)

Taking into account that , the equality no. (3) turns into

then in .  (4)

From and (4) we get and (1) leads to .

From and the fact that the points E, B, C belong to the circle with the center O and radius R, we deduce that the point M is identical with the center O and that [BC] is the diameter of the circle, which means that , that is the triangle ABC is right-angled in A.

THE ALTITUDE THEOREM

If [AD] is the altitude within the triangle ABC which is right-angled in A, then .

Demonstration. The triangle ABV which is right-angled in A is inscribed in the circle with the center O and diameter BC. We draw , and note as E the point of the intersection between the straight line AD and the circle; see the figure! As [BC] is a diameter perpendicular to the chord AE in the point D, we deduce that . According to the power of the interior point D towards the circle, we have (2)

Taking (1) into account, the equality (2) becomes .

THE CATHETUS THEOREM

If in the triangle ABC which is right-angled in A, A, then .

Demonstration. The triangle ABC being right-angled in A is inscribed in the circle with the center O and diameter AC; . As the triangle ABC is right-angled in A, the straight line BA is tangent to the circle in the point A. According to the power of the point B which is external to the circle, we have , that is .

Notice … In parallel, it shall be demonstrated that.

PYTHAGOREAN THEOREM GENERALISED

I. If the triangle ABC has the acute angle B and A, then .

Demonstration. The triangle ADC being right-angled in D is inscribed in the circle with the center O and diameter AC; The straight line BO intersects the circle in the points E and F; ; see the figure!

According to the power of the point B which is external to the circle, we have , resulting in , which turns into , then into

(1)

As [BO] is a median line in the triangle ABC and its length is given by , then the equality no. (1) becomes

,

leading us to .

II. If the triangle ABC has the obtuse angle B and A, then

Demonstration. The triangle ABD being right-angled in D is inscribed in the circle with the center O and diameter AB; . The straight line CO intersects the circle in the points E and F; ; see the figure!

According to the power of the point C which is external to the circle, we have , leading to , that is (1)

Knowing that [CO] is a median line in the triangle ABC and its length is given by , the equality no.(1) becomes

,

resulting in






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