Notes on Basic Euclidean Geometry

Ngoc-Loi Nguyen
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1. Euclidean Parallel Postulate

1.1. Methods of proof

Euclidean geometry is constructive in asserting the existence and uniqueness of certain geometric figures, and these assertions are of a constructive nature: that is, we are not only told that certain things exist, but are also given methods for creating them with no more than a compass and an unmarked straightedge.

Euclid often used proof by contradiction. Euclidean geometry also allows the method of superposition, in which a figure is transferred to another point in space.

1.2. Corollary 1

1.3. Corollary 2

2. Triangle Theorem

2.1. Triangle Theorem 1 for 1 same length : ASA

If and and . Note 2 angles at 2 ends of the equal side of triangle.

Then are congruent

2.1.1. Proof

There’s only 1 line parallel to AB from E, similarly only 1 line parallel to CA from F.

So these 2 triangles are congruent due to uniqueness property

2.2. Triangle Theorem 2 for 2 same length : SAS

If and . Note the angle made by 2 equal sides of triangle.

Then are congruent

2.2.1. Proof

There’s only 1 line from D to F to create this triangle.

So these 2 triangles are congruent due to uniqueness property

2.3. Triangle Theorem 3 for 3 same length : SSS

If and

Then are congruent

2.3.1. Proof

Draw a circle at E wirh radius ED and another one at F with radius FD, they cuts at single point D to create a unique..

So these 2 triangles are congruent due to uniqueness property

3. Isosceles Triangle Theorem (ITT)

If then

3.1. Proof

Let AD be bisector of

By SAS, we have

Hence

3.2. Corollary 1

If then AB = AC

3.3. Corollary 2

If AD is bisector of then

3.3.1. Proof

From the Proof 1 above

3.4. Corollary 3

If AD is bisector of BC then AD is also bisector of

3.4.1. Proof

From the Proof 1 above

4. Parallelogram

A parallelogram has 2 pairs of parallel sides

4.1. Parallelogram Theorem

A Parallelogram has 2 pairs of equal sides

4.1.1. Proof

We have by ASA so their corresponding sides must be equal

5. Basic Proportionality Theorem (Thales Theorem)

In a triangle, draw a line parallel to a side as in the figure above, we have

5.1. Proof

We will use the notation to denote the area of a triangle ABC

Since have the same base and altitude , we have

(1)

Since has the same altitude, we have

(2)

Similarly

(3)

Hence, by Eq(1)

5.2. Corollary 1

The Thales theorem can be expressed as

5.2.1. Proof

Draw EF parallel to AB, by Thales theorem, we have

6. Similar Triangles Theorem (AA)

Triangles are similar if they have 2 pairs of same angles then we have ratio of coressponding

6.1. Proof

If similar triangles have 2 pairs of equal angles, then the third pair must be equal, and we’re able to find a point D on AB and point E on AC to have . As Corollary, we have , so . By Thales theorem, we have

, but , then

7. Pythagoras theorem

For a right triangle, square of hypothenus equal sum of square each sides

7.1. Proof

Let AD be perpendicular to BC. We have , so, hence by Thales theorem we have

Similarly, we have , so

Sum up these 2 equations, we have

8. Circumcenter Theorem

All perpendicular bisectors of a triangle concur at one point called circumcenter as a center of its circumscribing circle.

8.1. Proof

Bisectors of AB and BC meet at O. By ITT, we have , hence there’s a circle circumscribing

8.2. Corollary 1

All 3 bisectors of 3 sides of a triangle concur at ONE point called circumcenter.

8.2.1. Proof

There’s a unique line from O to mid-point of AC. By ITT, it must perpendicular to AC.

8.3. Corollary 2

The midpoint of hypotenuse of a right triangle is the center of its circumscribed circle

8.3.1. Proof

Draw a circle with diameter as one side of a triangle whose 3^rd virtex in on the circle. Now we need to prove it is a right triangle.

By ITT, we have

As sum of all angles in a triangle is 180⁰, so

Hence it is a right triangle.

9. Ceva Theorem

Let ABC be a triangle, and let D, E, F be points on lines BC, CA, AB, respectively. Lines AD, BE, CF concur iff (if and only if)

(4)

9.1. Proof

Let X be the point those 3 lines concur at, and let be notation for area of a triangle ABC

We have

Similarly

And

Conversely, given Eq.(4) and let AD cuts BE at X, then let CX cuts AB at F*, we have AD, BE and CF* concur, so

(5)

Eqs.(4) & (5) give

10. Angular Bisector of Triangle

In a triangle ABC, let P on BC such that AP bisects , then

10.1. Proof

From B, draw a line parallel to AP and cuts AC at D, then so is isosceles, hence

Thales theorem give

11. Inner Cirle of Triangle

3 bisectors of angles in a triangle concur at a point as center of its inner circle.

11.1. Proof

3 bisectors AP, BQ and CR have

So they concur at a point, say O, by Ceva theorem. Let OI, OJ, OK be perpendicular to BC, CA, AB. By ASA, we have , similarly

12. Centroid Theorem

All bisects of triangle sides concur at one point called Centroid at 1/3 from the base.

12.1. Proof

Let L, M and N be midpoints of BC, AC and AB. By Ceva theorem, these 3 lines must concur at a point, say G. Thales theorem gives , so by a scale of 2 by ASA theorem. Thus and . Similarly, we have .

13. Orthocenter Theorem

All altitudes of a triangle meet at one point called orthocenter.

13.1. Proof

We have

Thus they concur by Ceva theorem

14. Euler Line

Let O, G, H be circumcenter, centroid, orthocenter of triangle ABC, then they are in line such that

14.1. Proof

Let O be circumcenter, M be midpoint of BC and G be centroid. Extend OG such that GH = 2 GO. Then we have due to SAS with scale of 2, hence , thus AH//OM. We have , ie AH is an altitude of BC. Similarly we have BH and CH be altitudes of AC and AB respectively.

OGH is known as Euler line