[PDF] Geometry - Theorems about triangles





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BASIC GEOMETRIC FORMULAS AND PROPERTIES

This handout is intended as a review of basic geometric formulas and Pythagorean Theorem (for right triangles only): ... BASIC PROBLEMS OF GEOMETRY.



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whose coordinates are given and to find the area of the triangle formed by three Solution : Let us apply the distance formula to find the distances PQ



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Geometry - Theorems about triangles

15-Dec-2013 as the three pairs of areas which we proved to be equal cancel. 4 Using the 1. 2 absin? formula for the area of a triangle we have.



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From the Figure we can deduce that we have been given 2 sides and the included angle. We can use the cosine formula to deduce the length of side a. a2. = b2 + 



Math Handbook of Formulas Processes and Tricks Geometry

Chapter 4. Triangles - Basic. Geometry. Length of Height Median and Angle Bisector. Height. The formula for the length of a height of a triangle is derived.



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  • Quelle sont les formule du triangle ?

    La formule de l'aire d'un triangle est : Aire d'un triangle = (Base × hauteur) : 2 soit : A = (B × h) : 2. Pour calculer l'aire d'un triangle rectangle, on peut utiliser la formule de l'aire d'un rectangle, mais il faudra diviser le résultat obtenu par 2.

  • Quelles sont les 4 droites remarquables d'un triangle ?

    Une droite est dite remarquable dans un triangle lorsqu'elle poss? une ou plusieurs propriétés quel que soit le triangle. Il existe 4 types de droites remarquables dans le triangle : la médiane, la médiatrice, la hauteur et la bissectrice.

  • Comment calculer le triangle ABC ?

    Donc l'aire du triangle ABC est donnée par : On a donc le résultat suivant : L'aire d'un triangle est égale au produit de la longueur d'un côté du triangle (base relative b) par sa hauteur h relative divisé par 2. Aire (ABC) = (base × hauteur) ÷ 2 = (b × h) ÷ 2.
  • Il existe quatre principaux types de triangles qui ont chacun des propriétés particulières : le triangle quelconque, le triangle isocèle, le triangle équilatéral et le triangle rectangle. Un triangle poss? trois côtés, trois sommets et trois angles. On le nomme par les lettres qui se trouvent à chacun de ses sommets.

Warm-up

Theorems about trianglesGeometry

Theorems about triangles

Misha Lavrov

ARML Practice 12/15/2013

Misha LavrovGeometry

Warm-up

Theorems about trianglesProblem

Solution

Warm-up problem

Lunes of Hippocrates

In the diagram below, the blue triangle is a right triangle with side lengths 3, 4, and 5.What is the total area of the green shaded regions?

Misha LavrovGeometry

Warm-up

Theorems about trianglesProblem

Solution

Solution

The lunes in the picture are formed by three semicircles whose

diameters are the three sides of the triangle.By the Pythagorean theorem, if we add the areas of the two small

semicircles, and subtract the area of the larger semicircle, we get 0. (In this case, the areas are 92

¼, 8¼, and252

But in the diagram, this is the difference between the green area and the blue area. So the green area is equal to the blue area, which we can compute to be 6.

Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

The angle bisector theorem

Suppose that in the triangle4ABC,ADis an angle bisector: \BADAE\CAD. ThenABAC

AEBDCD

.CABDI have three proofs of this theorem.

Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Angle bisector exercise

We are given a triangle with the following property: one of its angles is quadrisected (divided into four equal angles) by the

height, the angle bisector, and the median from that vertex.This property uniquely determines the triangle (up to scaling). Find

the measure of the quadrisected angle. (Hint: go wild with the angle bisector theorem.)

Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Solution

The base is partitioned into four segments in the ratio x:x:y:2xÅy. Suppose the length of the left-hand side of the triangle is 1. Then the length of the angle bisector is also 1. Applying the angle bisector theorem to the large triangle, we see that the length of the right-hand side is

2xÅ2y2xAE1Åyx

. But if we apply the angle bisector theorem to the left half of the triangle, we obtain

2xÅyy

AE1Å2xy

for the same length. Thereforeyx AE2xy , so x:yAE1:p2. Now apply the angle bisector theorem a third time to the right triangle formed by the altitude and the median. The segments in the base are in the ratiox:yAE1:p2, so the altitude and the median form the same ratio. As this is a right triangle, it must be a 45
±-45±-90±triangle. So the quadrisected angle is right.Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Stewart"s theorem

The line in the diagram below is no longer an angle bisector but just an arbitrary line. In addition to the labeled lengths, the base of the triangle has lengthaAEmÅn.bdcnmStewart"s Theorem.In this setting, b2mÅc2nAEa(d2Åmn). It"s conventional to memorize: "manÅdadAEbmbÅcnc", or "A man and his dad put a bomb in the sink."

Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Stewart"s theorem

Proof Draw the height,h, and label the unknown length in the base byx.bdhcnxm-xThen we have: 8>>< >:b

2AEh2Å(nÅx)2

c

2AEh2Å(m¡x)2

d

2AEh2Åx2.

Eliminate firsthand thenxto obtain Stewart"s theorem.Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Stewart"s theorem

Proof Draw the height,h, and label the unknown length in the base byx.bdhcnxm-xThen we have: 8>>< >:b

2AEh2Å(nÅx)2

c

2AEh2Å(m¡x)2

d

2AEh2Åx2.

Eliminate firsthand thenxto obtain Stewart"s theorem.Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Problems1Two sides of a triangle are 4 and 9; the median drawn to the

third side has length 6. Find the length of the third side.2A right triangle has legsaandband hypotenusec. Two

segments from the right angle to the hypotenuse are drawn, dividing it into three equal parts of lengthxAEc3 .bpqaxxxIf the segments have lengthpandq, prove thatp2Åq2AE5x2.Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Solutions1Ifais the length of the third side, thenmAEnAEa2 , and we have

8aÅ812

aAEaµ

36Å14

which yieldsa2AE50 oraAE5p2.

2Applied first topand then toq, Stewart"s theorem yields two

equations:(a2xÅb2(2x)AE3x(p2Å2x2), a

2(2x)Åb2xAE3x(q2Å2x2).

Adding these, we get(a2Åb2)(3x)AE(p2Åq2Å4x2)(3x), so p

2Åq2AEa2Åb2¡4x2. Buta2Åb2AEc2AE9x2, sop2Åq2AE5x2.Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Ceva"s theorem

In a triangle4ABC, letX,Y, andZbe points on the sides oppositeA,B, andC, respectively.CBAXYZTheAX,BY, andCZmeet at a single point if and only if: AZZB

¢BXXC

¢CYYA

AE1. (Terminology: we sayAX,BY,CZareconcurrent.)Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Ceva"s theorem

Proof (one direction)

We have

BXXC

AE[ABX][AXC].CBAXYZCBAXYZCBAXYZTherefore

AZZB

¢BXXC

¢CYYA

AE[ACO][BCO]¢[ABO][ACO]¢[BCO][ABO]AE1.Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Ceva"s theorem

Proof (one direction)

We have

BXXC AZZB

¢BXXC

¢CYYA

AE[ACO][BCO]¢[ABO][ACO]¢[BCO][ABO]AE1.Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Ceva"s theorem

Proof (one direction)

We have

BXXC AZZB

¢BXXC

¢CYYA

AE[ACO][BCO]¢[ABO][ACO]¢[BCO][ABO]AE1.Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Problems1Show that the following standard triples of lines are concurrent: the medians (easy); the angle bisectors (medium); the altitudes (hard).2A circle inscribed in4ABC(the incircle) is tangent toBCat

X, toACatY, toABatZ. Show thatAX,BY, andCZare

concurrent.3Three squares are drawn on the sides of4ABC(i.e. the square onABhasABas one of its sides and lies outside

4ABC). Show that the lines drawn from the verticesA,B,C

to the centers of the opposite squares are concurrent.4Prove that for any pointsX,Y,ZonBC,AC,AB, AZZB

¢BXXC

¢CYYA

AEsin\ACZsin\ZCB¢sin\BAXsin\XAC¢sin\CBYsin\YBA.Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Solutions11For the medians,

AZZB

AEBXXC

AECYYA

AE1, so their product is 1.2For the angle bisectors, use the angle bisector theorem: AZZB

¢BXXC

¢CYYA

AEACBC

¢ABAC

¢BCAB

AE1.3For the altitudes,4ABXand4CBZare similar, because \ABXAE\CBZAE\ABCand\AXBAE\CZBAE90±. Therefore BZBX

AEBCAB

, which lets us simplify the Ceva"s theorem product in the same way as above, after rearranging.2AYAEAZbecause these are the two tangent lines fromAto the incircle. Similarly,BXAEBZandCXAECY, and the result follows.

Misha LavrovGeometry

Warm-up

Theorems about trianglesThe angle bisector theorem

Stewart"s theorem

Ceva"s theorem

Solutions3LetA0,B0,C0be the centers of the squares oppositeA,B,C.

Then4ABA0and4CBC0have the same area:

\ABA0AE\CBC0AE\ABCÅ45±, andBA0:BC0AECB:AB.

In the proof of Ceva"s theorem we had

BXXC

AE[ABX][ACX], and by the

same argument we have BXXC

AE[ABA0][ACA0]. Therefore

AZZB

¢BXXC

¢CYYA

as the three pairs of areas which we proved to be equal cancel.4Using the 12 absinµformula for the area of a triangle, we have BXXC

AE[AXB][AXC]12

AC¢AX¢sin\XAC1

2

AB¢AX¢sin\BAXAEACAB

¢sin\XACsin\BAX.

Doing the same for all three ratios yields the formula we want.

Misha LavrovGeometry

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