In this article, we will be looking at some of the basic theorems of straight lines and triangles. In most of the Geometry based questions in competitive tests, theorems offer the starting point of the solution. Over the next few articles, we will be exploring all the theorems regarding lines, triangles, and circles that one needs to know from CAT preparation point of view.

**Theorem 6: If three sides of one triangle are respectively equal to the three sides of another triangle, then the two triangles are congruent.**

This is also known as SSS theorem.

For example, if AB = PQ, BC = QR, and AC = PR, then by SSS theorem, triangles ABC and PQR are said to be congruent.

**Theorem 7: If in two right angled triangles, the hypotenuse and a side of on triangle are respectively equal to the hypotenuse and a side of the other, then the two triangles are congruent.**

In triangles ABC and PQR, let angle B and Q be 90º.

If AC = PR and BC = QR, then the two triangles are congruent. This is quite simple to understand as in both the triangles, by Pythagoras theorem, square of the hypotenuse = sum of the squares of the other two sides. Hence, by both SSS and SAS tests, we will get the triangles to be congruent. This is also known as the RHS theorem.

**Theorem 8: If two sides of a triangle are not equal, then the greater side has the greater angle opposite to it.**

In triangle ABC, if AC > BC > AB. Therefore, ∠ ABC > ∠ BAC > ∠ ACB.

Whenever AC > AB, ∠ B > ∠ C.

This also gives us another theorem: If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.

Another point to be noted: In a right angled triangle, the two angles other than the right angle are acute and hence the hypotenuse is the largest side.

**Theorem 9: The locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment joining the two points.**

If B and C are two points and let D be the midpoint of BC. (DB = DC)

If A perpendicular is drawn to line BC from point A, triangles ADB and ADC can be proved to be congruent. Hence, any point on the perpendicular bisector of BC will be equidistant from B and C. Thus, the locus of A which is equidistant from B and C is the perpendicular bisector of the line segment BC.

**Theorem 10: Suppose two straight lines are cut by a transversal and any one of the following three conditions hold, namely,**

**a pair of alternate angles are equal****a pair of corresponding angles are equal****a pair of interior angles on the same side of the transversal add up to 180º**

**then the two straight lines are parallel.**

Using this theorem, if p is transversal and l and m are two straight lines, if any one of the following three conditions hold, namely,

- a pair of alternate angles are equal (say ∠ C and ∠ F)
- a pair of corresponding angles are equal (say ∠ B and ∠ F)
- a pair of interior angles on the same side of the transversal add up to 180º (say ∠ D and ∠ F)

then, these two straight lines are parallel. This also means that if two parallel straight lines l and m are cut by a transversal, then the alternate angles are equal, the corresponding angles are equal, and the sum of the interior angles on the same side of the transversal is equal to 180º.

∠ A = ∠ D = ∠ E = ∠ H

∠ B = ∠ C = ∠ F = ∠ G

∠ C + ∠ E = ∠ D + ∠ F = 180º

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