Exercise 5.2: Introduction to Euclid's Geometry — NCERT Solutions

Euclid's fifth postulate is complicated in its original form. A simpler equivalent statement is Playfair's Axiom:

'For every line $\ell$ and for every point $P$ not lying on $\ell$, there exists a unique line through $P$ that is parallel to $\ell$.'

In other words, through a point outside a given line, exactly one parallel line can be drawn. This version is logically equivalent to Euclid's fifth postulate but far easier to state and use.

Since $C$ lies between $A$ and $B$, by the betweenness property: $AC + BC = AB.$

We are given that $AC = BC$. Substituting $BC = AC$: $AC + AC = AB$ $2\,AC = AB.$

Using Euclid's axiom (equals divided by equals are equal), divide both sides by $2$: $AC = \tfrac{1}{2}\,AB.$

Hence proved. Such a point $C$ is called the midpoint of $AB$.

Let $AB$ be a line segment and suppose it has two midpoints $C$ and $D$.

Since $C$ is a midpoint: $AC = \tfrac{1}{2}\,AB. \quad (1)$ Since $D$ is a midpoint: $AD = \tfrac{1}{2}\,AB. \quad (2)$

From (1) and (2), by Euclid's axiom 'things equal to the same thing are equal to one another': $AC = AD.$

But both $C$ and $D$ lie on the same segment $AB$ and are measured the same distance from $A$. Two points at the same distance from $A$ along $AB$ must coincide, so: $C = D.$

This contradicts our assumption of two distinct midpoints. Hence a line segment has one and only one midpoint.

Since $B$ lies between $A$ and $C$, the segment $AC$ is made up of two non-overlapping parts $AB$ and $BC$.

By Euclid's common notion 'the whole is equal to the sum of its parts', the whole segment $AC$ equals the sum of its parts $AB$ and $BC$: $AB + BC = AC.$

Hence proved.

The points lie in the order $A, B, C, D$ on a line.

Given: $AB = CD. \quad (1)$

Add $BC$ to both sides (Euclid's axiom: if equals are added to equals, the wholes are equal): $AB + BC = CD + BC. \quad (2)$

Now use the betweenness/whole-is-sum-of-parts property:

• $AB + BC = AC$ (since $B$ lies between $A$ and $C$),
• $CD + BC = BC + CD = BD$ (since $C$ lies between $B$ and $D$).

Substituting into (2): $AC = BD.$

Hence proved.

Statement: Two distinct lines cannot have more than one point in common.

Justification (by contradiction): Suppose two distinct lines $\ell$ and $m$ have two common points, say $P$ and $Q$.

Then both $\ell$ and $m$ pass through the two distinct points $P$ and $Q$. But Euclid's first postulate says that through two distinct points there passes exactly one line. This means $\ell$ and $m$ must be the same line — contradicting that they are distinct.

Hence two distinct lines can have at most one point in common.