Question

From an external point A , two tangents AB and AC are drawn to the circle with center O . Then OA is the dash bisector of BC

Answer 1

Answer:

perpendicular

Step-by-step explanation:

as BC is a chord of the circle and the radius is the perpendicular bisector of any chord in the circle

Answer 2

Concept:

A perpendicular bisector is a line that bisects another line segment at a right angle, through the intersection point.

Tangents on a circle from an external point are equal in length.

When 2 lines of equal length are inclined with the same line their angle is the same.

A Congruent triangle means the same size and shape. Types of congruency:

1. SSS: When all three sides are equal to each other on both triangles, the triangle is congruent

2. AAS: If two angles and a non-included (you can think of it as outside) side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.

3. ASA: If two angles and the included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.

4. SAS: If any two angles and the included side are the same in both triangles, then the triangles are congruent.

Knowledge of Corresponding Parts of Congruent Triangles.

Given:

A is an external point from where two tangents AB and AC are drawn from the circle with center O. OA is a bisector of BC.

To find:

We need to find the type of bisector is OA to BC.

Solution:

With the given information we can draw the below diagram,

Let OC intersect AB at F

Therefore,

ΔCFA and ΔCFB

We know that,

CA=CB

∠ACF=∠BCF

and CF=CF               (common)

∴ΔCFA=ΔCFB          (by SAS congrunce criteria)

Therefore,

⇒AC=BC

and ∠AFC=∠BFC

By using CPCT

∠AFC+∠BFC=180°       (angle of a line is 180°)

⇒2∠AFC=180°

⇒∠AFC=∠BFC=$\frac{180}{2}$=90°

Therefore, we can conclude that OA is the perpendicular bisector.