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**(1) Line Segment :**Â **It is a line with two end points.**It is denoted by .

**(2) Ray:Â ****It is a line with one end point.
**It is denoted by Â .

**(3) Collinear Points : If three or more points lie on the same line, they are called collinear points; otherwise they are called non â€“ collinear points.**For the figure shown above, A, B and C are collinear points.For the figure shown above, A, B and C are non - collinear points.

**(4) Angle : It is formed when two rays originate from the same end point. The rays which form an angle are called its arms and the end point is called the vertex of the angle.**

**(5) Types of Angles:
**

**(ii) Right angle:Â **It is the angle whose measure is equal to 90áµ’.

**(iii) Obtuse angle:Â **It is the angle whose measure is greater 90áµ’ than but less than 180áµ’.

**(iv) Straight angle:Â **It is the angle whose measure is equal to 180áµ’.

**(v) Reflex angle:Â **It is the angle whose measure is greater 180áµ’ than but less than 360áµ’.

**(vi) Complementary angles:Â **The two angles whose sum is 90áµ’ are known as complementary angles.For the figure shown above, the sum of angles a & b is 90áµ’, hence these two angles are complementary angles.

**(vii) Supplementary angles:Â **The two angles whose sum is 180áµ’ are known as supplementary angles.For the figure shown above, sum of the angles 45áµ’ and 135áµ’ is 180áµ’, hence these two angles are supplementary angles.

**(viii) Adjacent angles:Â **Two angles are said to be adjacent if they have a common vertex, a common arm and their non-common arms are on different side of the common arm.

When two angles are adjacent, then their sum is always equal to the angle formed by the two non common arms.For the figure shown above, âˆ ABD and âˆ DBC are adjacent angles. Here, ray BD is the common arm and B is the common vertex. And ray BA and BC are non common arms.

Here, âˆ ABC = âˆ ABD + âˆ DBC.

**(ix) Linear pair of angles:Â **Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. The linear pair of angles must add up to 180áµ’.For the figure shown above, âˆ ABD and âˆ DBC are called linear pair of angles.

**(x) Vertically Opposite angles:Â **These are the angles opposite each other when two lines cross.For the figure shown above, âˆ AOD and âˆ BOC are vertically opposite angles. Also, âˆ AOC and âˆ BOD are vertically opposite angles.

** (xi) Intersecting lines:Â **These are the lines which cross each other.For the figure shown above, lines PQ and RS are the intersecting lines.

**(xii) Non-intersecting lines:Â **These are the lines which do not cross each other.For the figure shown above, lines PQ and RS are the non-intersecting lines.

**(6) Pair of Angles:
**

Here, the pair of vertically opposite angles formed are (i) âˆ AOC and âˆ BOD (ii) âˆ AOD and âˆ BOCÂ And we need to prove that âˆ AOC = âˆ BOD and âˆ AOD = âˆ BOC.

Here, ray OA stands on line CD. Hence, âˆ AOC + âˆ AOD = 180Â° as per linear pair axiom. Similarly, âˆ AOD + âˆ BOD = 180Â°.

On equating both, we get, âˆ AOC + âˆ AOD = âˆ AOD + âˆ BOD

Thus, âˆ AOC = âˆ BOD

Similarly, it can be proved that âˆ AOD = âˆ BOC.

**(7) Some Examples:
**

Hence, âˆ PQS +âˆ PQR = 180Â° i.e. âˆ PQS = 180Â° - âˆ PQR - (i)

Also, from the figure, we can see that âˆ PRQ and âˆ PRT forms a linear pair.

Hence, âˆ PRQ +âˆ PRT = 180Â° i.e. âˆ PRT = 180Â° - âˆ PRQ

Given, âˆ PQR = âˆ PRQ

Therefore, âˆ PRT = 180Â° - âˆ PQR - (ii)

From (i) and (ii),

âˆ PQS = âˆ PRT = 180Â° - âˆ PQR

Thus, âˆ PQS = âˆ PRT

** For Example:** OP, OQ, OR and OS are four rays. Prove that âˆ POQ + âˆ QOR + âˆ SOR + âˆ POS = 360Â°.Firstly, let us make ray OT as shown in figure below to make a line TOQ.From the above figure, we can see that, ray OP stands on line TOQ.

Hence, as per linear pair axiom, âˆ TOP + âˆ POQ = 180Â° - (i)

Similarly, from the figure, we can see that, ray OS stands on line TOQ.

Hence, as per linear pair axiom, âˆ TOS + âˆ SOQ = 180Â°

But, from the figure, âˆ SOQ = âˆ SOR + âˆ QOR

So, âˆ TOS + âˆ SOR + âˆ QOR = 180Â° - (ii)

On adding (i) & (ii), we get,

âˆ TOP + âˆ POQ + âˆ TOS + âˆ SOR + âˆ QOR = 360Â°

From the figure, âˆ TOP + âˆ TOS = âˆ POS

Therefore, âˆ POQ + âˆ QOR + âˆ SOR + âˆ POS = 360Â°.

**(8) Parallel lines and a Transversal:
**

Here, âˆ 1, âˆ 2, âˆ 7 and âˆ 8 are exterior angles.

Here, âˆ 3, âˆ 4, âˆ 5 and âˆ 6 are interior angles.

Here, (i) âˆ 1 and âˆ 5 (ii) âˆ 2 and âˆ 6 (iii) âˆ 4 and âˆ 8 (iv) âˆ 3 and âˆ 7 are corresponding angles.

Here, (i) âˆ 4 and âˆ 6 (ii) âˆ 3 and âˆ 5 areÂ alternate interior angles.

Here, (i) âˆ 1 and âˆ 7 (ii) âˆ 2 and âˆ 8 areÂ alternate exterior angles.

** Axiom 1:** If a transversal intersects two parallel lines, then each pair of corresponding angles is equal.

**(9) Lines Parallel to the Same Line:
**

** For Example:Â **If AB || CD, EF âŠ¥ CD and âˆ GED = 126Â°, find âˆ AGE, âˆ GEF and âˆ FGE.From the figure, we can see that, âˆ AGE and âˆ GED forms alternate interior angles.

Therefore, âˆ AGE = âˆ GED = 126Â°

From the figure, we can see that, âˆ GEF = âˆ GED - âˆ FED = 126Â° - 90Â° = 36Â°

Again from the figure, we can see that, âˆ FGE and âˆ AGE forms linear pair.

Therefore, âˆ FGE + âˆ AGE = 180Â°

âˆ FGE = 180Â° - 126Â° = 54Â°.

** For Example:** AB || CD and CD || EF. Also, EA âŠ¥ AB. If âˆ BEF = 55Â°, find the values of x, y and z.From the figure, we can see that, âˆ y and âˆ DEF forms interior angles on the same side of the transversal ED.

Therefore, y + 55Â° = 180Â° => y = 180Â° - 55Â° = 125Â°

From the figure, we can see that, AB || CD, so as per corresponding angles axiom x = y.

So, x = 125Â°

From the figure, we can see that, AB || CD and CD || EF, hence, AB || EF.

Therefore, âˆ EAB + âˆ FEA = 180Â° - (i)

From the figure, âˆ FEA = âˆ FEB + âˆ BEA.

Substituting in (i), we get,

âˆ EAB + âˆ FEB + âˆ BEA = 180Â°

90Â° + z + 55Â° = 180Â° i.e. z = 35Â°.

**(10) Angle Sum Property of a Triangle:
**

*Proof*:

For the given triangle PQR, we need to prove that âˆ 1 + âˆ 2 + âˆ 3 = 180Â°.

Firstly, we will draw line XPY parallel to QR passing through P as shown in figure below.From the figure, we can see that âˆ 4 + âˆ 1 + âˆ 5 = 180Â° - (1)

Here, XPY || QR and PQ, PR are transversals. So, âˆ 4 = âˆ 2 and âˆ 5 = âˆ 3 (Pairs of alternate angles).

Substituting âˆ 4 and âˆ 5 in (1), we get, âˆ 1 + âˆ 2 + âˆ 3 = 180Â°.

Hence, the sum of the angles of a triangle is 180Â°.

** For Example:Â **The side QR of âˆ† PQR is produced to a point S. If the bisectors of âˆ PQR and âˆ PRS meet at point T, then prove that âˆ QTR = 1/2 âˆ QPR.We know that, the exterior angle of triangle is equal to the sum of the two interior angles.

So, âˆ TRS = âˆ TQR + âˆ QTR i.e. âˆ QTR = âˆ TRS - âˆ TQR â€“ (i)

Similarly, âˆ SRP = âˆ QPR + âˆ PQR â€“ (ii)

From the figure, âˆ SRP = 2 âˆ TRS and âˆ PQR = 2 âˆ TQR

Hence, equation (ii) becomes,

2 âˆ TRS = âˆ QPR + 2 âˆ TQR

âˆ QPR = 2 âˆ TRS - 2 âˆ TQR => Â½ âˆ QPR = âˆ TRS - âˆ TQR â€“ (iii)

On equating (i) and (iii), we get,

âˆ QTR = Â½ âˆ QPR.

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