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Angle and Angle Bisector

Lines and Angles III - Concepts

Angle Sum Property of a Triangle

Exterior Angle Property

Angle and Angle Bisector

Class - 9th CBSE Subjects

Concept Explanation

Angle and Angle Bisector

Angle: The corners made by the intersection of two lines or line segments are called Angles.

Angle Bisector: A ray AX is said to be the bisector of $angle BAC$ , if X is a point in the interior of $angle BAC$ , and $angle BAX=angle XAC.$

Thus, if ray AX is the bisector of $angle BAC$ , then

$angle BAX=angle CAX=frac{1}{2}angle BAC$

Theorem: 1 If two parallel lines are intersected by a transversal, the bisectors of any pair of alternate interior angles are parallel.

Converse of theorem: If the bisectors of a pair of alternate angles formed by a transversal with two given lines are parallel, then the given lines are parallel.

Theorem 2: If two parallel lines are intersected by a transversal, then bisectors of any two corresponding angles are parallel.

Converse of theorem: If the bisectors of any pair of corresponding angles formed by a transversal with two given lines are parallel, then the given lines are parallel.

Theorem 3: If the bisectors of angles $angle ABC;and;angle ACB$ of a triangle ABC meet at a point O, then

$angle BOC =90^{0}+frac{1}{2}angle A$

Proof:

Given: A triangle ABC such that the bisectors of $angle ABC;and;angle ACB$ meet at a point O.

To Prove:

$angle BOC =90^{0}+frac{1}{2}angle A$

Proof: In BOC, we have

$angle 1+angle 2+angle BOC=180^{0}$

In triangle ABC

$angle A+angle B+angle C=180^{0}$ ................(i)

$Rightarrow angle A+2angle 1+2angle 2=180^{0}$ [ $because BO;and;CO$ are bisectors of $angle ABC$ and $angle ACB$ respectively. $therefore angle B=2angle 1;and;angle C=2angle 2$ ]

$Rightarrow frac{angle A}{2}+angle 1+angle 2=90^{0}$ [dividing both sides by 2]

$angle 1+angle 2=90^{0}-frac{angle A}{2}$ ................(ii)

Substituting the value from equation (ii) in equation (i), we get

$90^{0}-frac{angle A}{2}+angle BOC=180^{0}$

$angle BOC=180^{0}-90^{0}+frac{angle A}{2}$

$Rightarrow angle BOC=90^{0}+frac{angle A}{2}$

Hence the proof.

Theorem 4: If two parallel lines are intersected by a transversal, then the bisectors of the two pairs of interior angles enclose a rectangle.

Theorem 5: The sides AB and AC of a $bigtriangleup ABC$ are produced to P and Q respectively. If the bisectors of $angle PBC;and;angle QCB$ intersects at O, then

$angle BOC=90^{0}-frac{1}{2}angle A$

Sample Questions

(More Questions for each concept available in Login)

Question : 1

In $bigtriangleup ABC,angle B=45^{0},angle C=55^{0}$ and bisector of $angle A$ meets BC at a point D. Find $angle BAD$ .
A. $70^{0}$	B. $80^{0}$	C. $40^{0}$	D. $60^{0}$
Right Option : C
View Explanation

Question : 2

TQ and TR are the bisectors of $angle Q$ and $angle R$ respectively.If $angle QPR=80^{0}$ and $angle PRT=30^{0}$ Find $angle QTR$
A. $30^{0}$	B. $50^{0}$	C. $130^{0}$	D. $150^{0}$
Right Option : C
View Explanation