This document proves that if two chords AB and AC of a circle have their bisector of the angle BAC as a diameter, then the chords AB and AC are equal in length. It shows that triangles AMO and ANO are congruent since they share an angle and are on the diameter, meaning their bases AM and AN are also equal. Since the perpendicular from the center cuts chords in half, this means that AB and AC must be equal.

1. WELCOME

2. CIRCLE

3. A circle can be displaced as the locus of a point
which moves at a specifics distance from a
specific distance from a specific point

4. AB and ac are two chords of a circle and
the bisector of ˂BAC is a diameter of the
circle. Prove that AB=AC

5. B
C
DA
N
M
O
AB and AC are the equal chords. This chords join with point
A. AD is the bisector of ˂ BAC . O be the center of circle.
Therefore, ˂AMO = ˂ ANO = 90˚
˂MAO = ˂NAO ( Since AD is the bisector of ˂ BAC )

6. Therefore, ˂AOM = ˂ AON. Why.
(If two angles of two triangles are
equal , then the third angle also will
be equal then the third angle also will
be equal.)

7. Therefore AMO and ANO are congruent
(one side and two angles on that side of a
triangle are equal to one side and two angles
on that side of other triangle)

8. AM=AN (Corresponding sides of
congruent triangles)
Since the perpendicular at from the
centre of the circle to the chord passes
through the midpoint of the chord.
Therefore M is the midpoint of AB and
N is the midpoint of AC

9. Since AM=AN
½ AB =½AC
i.e., AB =AC

10. THANKYOU