Question

Complete the following activity.
Theorem : A perpendicular drawn from the centre of a circle on its chord bisects the
Given : seg AB is a chord of a circle with centre o
seg OPI chord AB
To prove : seg AP = seg
Proof Draw seg OA and seg OB
In AOPA and A
ZOPA ZOPB
seg OP I chord AB
common side
hypotenuse OA = hypotenuse OB
Fig. 6.4
...radii of the same circle
', ΔΟΡΑΣ Δ
hypotenuse side theorem
seg PA seg PB
,
seg OP,my exam is going help me out​

Answer 1

To prove that the perpendicular from the centre to a chord bisect the chord.

Consider a circle with centre at O and AB is a chord such that OX perpendicular to AB

To prove that   AX=BX

In ΔOAX and ΔOBX

∠OXA=∠OXB  [both are 90 ]

OA=OB  (Both  are radius of circle )

OX=OX  (common side )

ΔOAX≅ΔOBX

AX=BX  (by property of congruent triangles )

hence proved.