As every the Angle Bisector theorem, the angle bisector of a triangle bisects the opposite next in together a means that the proportion of the two line-segments is proportional to the ratio of the other two sides. Thus the relative lengths the the opposite next (divided by angle bisector) are equated to the lengths the the various other two sides of the triangle. Angle bisector theorem is applicable to all varieties of triangles. 

Class 10 students have the right to read the ide of angle bisector to organize here along with the proof. Personally from edge bisector theore, we will also discuss right here the external angle theorem, perpendicular bisector theorem, converse of angle bisector theorem.

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Table that Contents:

What is edge Bisector Theorem?

An edge bisector is a straight line attracted from the peak of a triangle come its opposite next in such a way, that it divides the angle right into two equal or congruent angles. Now let united state see, what is the edge bisector theorem.

According come the angle bisector theorem, the edge bisector of a triangle divides the opposite side right into two components that space proportional to the other two sides of the triangle.

Interior angle Bisector Theorem

In the triangle ABC, the edge bisector intersects next BC at point D. See the number below.


As every the edge bisector theorem, the ratio of the heat segment BD to DC equates to the ratio of the length of the side abdominal to AC.

(fracleft DC ight =fracleft left )

Conversely, as soon as a allude D on the side BC divides BC in a ratio similar to the sides AC and AB, then the angle bisector that ∠ A is AD. Hence, according to the theorem, if D lies on the side BC, then,

(fracleft =fracleft left )

If D is exterior to the next BC, command angles and directed line segments are forced to be applied in the calculation.

Angle bisector to organize is applied when side lengths and also angle bisectors are known.

Proof of edge bisector theorem

We can easily prove the edge bisector theorem, by utilizing trigonometry here. In triangle ABD and also ACD (in the above figure) using regulation of sines, we can write;

(fracABBD=fracsinangle BDAsinangle BAD) ….(1)

(fracACDC=fracsinangle ADCsinangle DAC) ….(2)

The angles ∠ ADC and also ∠ BDA do a straight pair and also hence called adjacent supplementary angles. 

Since the sine that supplementary angles space equal, therefore,

Sin ∠ BDA = Sin ∠ ADC …..(3)


∠ DAC = ∠ bad (AD is the edge bisector)


Sin ∠ BDA = Sin ∠ ADC …(4)

Hence, from equation 3 and also 4, we have the right to say, the RHS that equation 1 and also 2 are equal, therefore, LHS will also be equal.

(fracleft left =frac AC ight )

Hence, angle bisector organize is proved. 


If the angles ∠ DAC and ∠ BAD room not equal, the equation 1 and equation 2 deserve to be written as:

(frac ab ight left ) sin ∠ poor = sin∠ BDA

(frac) sin ∠ DAC = sin∠ ADC

Angles ∠ ADC and ∠ BDA space supplementary, thus the RHS of the equations room still equal. Hence, us get

(frac ab ight ) sin ∠BAD = (frac AC ight ) sin ∠DAC

This rearranges to generalized view of the theorem.

Converse of angle Bisector Theorem

In a triangle, if the interior allude is equidistant from the 2 sides of a triangle then that allude lies ~ above the angle bisector that the angle developed by the 2 line segments.

Triangle edge Bisector Theorem


Extend the side CA to fulfill BE to fulfill at allude E, such that BE//AD.

Now we have the right to write,

CD/DB = CA/AE (since AD//BE) —-(1)

∠4 = ∠1

∠1 = ∠2

∠2 = ∠3

∠3 = ∠4

ΔABE is one isosceles triangle through AE=AB 

Now if we replace AE by abdominal in equation 1, us get;


Hence proved.

Perpendicular Bisector Theorem

According come this theorem, if a allude is equidistant indigenous the endpoints that a line segment in a triangle, then it is on the perpendicular bisector the the heat segment. 

Alternatively, we can say, the perpendicular bisector bisects the offered line segment into two equal parts, come which it is perpendicular. In case of triangle, if a perpendicular bisector is attracted from the vertex to the the opposite side, then it divides the segment right into two congruent segments.


In the over figure, the heat segment SI is the perpendicular bisector of WM.

External angle Bisector Theorem

The external angle bisector the a triangle divides the opposite next externally in the ratio of the political parties containing the angle. This condition occurs normally in non-equilateral triangles.


Given : In ΔABC, advertisement is the exterior bisector that ∠BAC and intersects BC developed at D. 

To prove : BD/DC = AB/AC

Constt: draw CE ∥ DA meeting abdominal at E


Since, CE ∥ DA and AC is a transversal, therefore, 

∠ECA = ∠CAD (alternate angles) ……(1)

Again, CE ∥ DA and also BP is a transversal, therefore,

∠CEA = ∠DAP (corresponding angles) —–(2)

But ad is the bisector the ∠CAP, 

∠CAD = ∠DAP —–(3)

As us know, political parties opposite to same angles room equal, therefore,


In ΔBDA, EC ∥ AD. 


AE = AC, 


Hence, proved.

Solved examples on edge Bisector Theorem

Go through the following instances to know the principle of the edge bisector theorem.

Example 1:

Find the worth of x for the provided triangle utilizing the edge bisector theorem.



Given that,

AD = 12, AC = 18, BC=24, DB = x

According to edge bisector theorem, 


Now instead of the values, we get

12/18 = x/24

X = (⅔)24

x = 2(8)

x= 16

Hence, the worth of x is 16.

Example 2:

ABCD is a quadrilateral in i beg your pardon the bisectors of angle B and angle D intersects ~ above AC at allude E. Present that AB/BC = AD/DC



From the provided figure, the segment DE is the angle bisector of edge D and BE is the interior angle bisector of angle B.

Hence, the using interior angle bisector theorem, us get

AE/EC = AD/DC ….(1)


AE/EC = AB/BC ….(2)

From equations (1) and (2), us get


Hence, AB/BC = AD/DC is proved.

Example 3.

In a triangle, AE is the bisector the the exterior ∠CAD the meets BC at E. If the value of abdominal = 10 cm, AC = 6 cm and also BC = 12 cm, discover the value of CE.


Given : ab = 10 cm, AC = 6 cm and also BC = 12 cm

Let CE is same to x.

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By exterior edge bisector theorem, we recognize that,

BE / CE = ab / AC

(12 + x) / x = 10 / 6

6( 12 + x ) = 10 x < by overcome multiplication>

72 + 6x = 10x

72 = 10x – 6x

72 = 4x

x = 72/4

x = 18

CE = 18 cm

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