Introduction to Euclid's Geometry: Axioms and Postulates - Apply Euclidean definitions, axioms, and postulates in geometric arguments

If a point $C$ lies between two points $A$ and $B$ such that $AC = BC$, then prove that $AC = \frac{1}{2} AB$. Explain by drawing the figure.

1. We are given $AC = BC$.
2. According to Euclid's axiom, 'if equals are added to equals, the wholes are equal'.
3. Let us add $AC$ to both sides of the given equation: $AC + AC = BC + AC$.
4. This gives $2AC = BC + AC$.
5. From the visual representation of the line segment, the point $C$ lies between $A$ and $B$, so $AC + BC$ coincides with the line segment $AB$.
6. According to the axiom 'things which coincide with one another are equal to one another', we have $AC + BC = AB$.
7. Therefore, $2AC = AB$.
8. Dividing both sides by $2$ (or taking halves), we get $AC = \frac{1}{2} AB$.

This solution utilizes Euclid's Axiom 2 (adding equals to equals) and Axiom 4 (coinciding objects) to establish a relationship between a part of a segment and the whole segment.

Prove that an equilateral triangle can be constructed on any given line segment.

1. Let $AB$ be the given line segment.
2. Using Euclid's Postulate 3 (a circle can be drawn with any centre and radius), draw a circle with centre $A$ and radius $AB$.
3. Similarly, draw another circle with centre $B$ and radius $BA$.
4. Let these two circles intersect at a point $C$.
5. Draw line segments $AC$ and $BC$ (Postulate 1).
6. Now, $AB = AC$ (radii of the same circle).
7. Also, $AB = BC$ (radii of the same circle).
8. According to Euclid's Axiom 1 (things equal to the same thing are equal to each other), since $AC = AB$ and $BC = AB$, then $AC = BC$.
9. Thus, $AB = BC = AC$. Since all three sides are equal, $\triangle ABC$ is an equilateral triangle.

This proof demonstrates the application of Euclid's Postulates 1 and 3 along with Axiom 1 to construct a geometric figure based on fundamental assumptions.

1. Given: $AC = BD$.
2. From the figure, we see that $AC = AB + BC$ and $BD = BC + CD$.
3. Substituting these into the given equation: $AB + BC = BC + CD$.
4. According to Euclid's Axiom 3 (If equals are subtracted from equals, the remainders are equal), we subtract $BC$ from both sides.
5. $(AB + BC) - BC = (BC + CD) - BC$.
6. Therefore, $AB = CD$.

This argument applies the axiom that subtracting the same quantity ($BC$) from equal total lengths ($AC$ and $BD$) maintains equality.

1. Let there be two distinct lines $l$ and $m$.
2. Suppose they have two distinct common points, $P$ and $Q$.
3. According to Euclid's Postulate 1, there is a unique line that passes through two distinct points.
4. This means both lines $l$ and $m$ must be the same line because they both pass through $P$ and $Q$.
5. This contradicts our assumption that lines $l$ and $m$ are distinct.
6. Hence, two distinct lines can have only one point in common.

This proof uses the method of contradiction based on the postulate that two points uniquely determine a line.