An edge bisector is identified as a beam that divides a given angle right into two angles v equal measures. The word bisector or bisection means dividing one thing right into two equal parts. In geometry, we generally divide a triangle and an angle by a line or ray which is considered as an edge bisector.

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2. | How to build an edge Bisector? |

3. | Angle Bisector Theorem |

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In geometry, us read around many types of angles. Every angle has its own property. Bisecting an angle gives a particular angle various characteristics. Following are the two major properties that an edge bisector holds.

1) Any point on the bisector the an angle is equidistant from the political parties of the angle.

2) In a triangle, the angle bisector divides the opposite next in the ratio of the surrounding sides.

Let"s shot constructing the edge bisector for an angle. In this section, us will view the actions to be followed for angle bisector construction. Let"s begin!

**Steps to construct an angle Bisector**

**Step 1:** Draw any angle, say ∠ABC.

**Step 2:** Taking B together the center and any kind of appropriate radius, draw an arc to crossing the beam BA and also BC at, say, E and also D respectively. (Refer come the figure below)

**Step 3:** Now, acquisition D and also E as centers and with a radius an ext than half of DE, attract an arc to crossing each other at F.

**Step 4:** draw ray BF. This ray BF is the compelled angle bisector of angle ABC.

## Angle Bisector Theorem

Let"s now know in detail critical property of the edge bisector that a triangle as proclaimed in the ahead section and then settle angle bisector problems. This residential or commercial property is well-known as the angle Bisector home of Triangle.

**Statement: An edge bisector of a triangle divides the opposite side into two segment that space proportional to the various other two political parties of the triangle.**

**Given :** Δ ABC; ad bisects ∠BAC

**To Prove:** AB/AC=BD/DC. We will prove this an outcome using properties of parallel lines and similarity of triangles.

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**Construction**: We build an assistant line passing v C and also parallel to side AB, intersecting advertisement extended at suggest E.

**Proof:**

**Step 1:**As AB||CE, ∠1=∠3 (∵ alternate angles)

**Step 2:**But, ∠1 = ∠2 (∵ AD is angle bisector)

**Step 4:**Now, ΔABD∼ΔECD (ByAA test of similarity) ∴AB/EC = BD/CD (corresponding edge of comparable triangles)(corresponding angle of similar triangles) p/y = a/b

**Step 5:**So from action 4, we obtained p/y = a/b. Likewise from action 3, we have x = y ⇒ p/x = a/b i.e., AB/AC = BD/DC

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