Question

2. A perpendicular drawn from the centre of a circle on its chord bisects the chord.
Complete the following activity to prove it.
Given: Seg AB is a chord of a circle with centre O.
seg OP chord AB.
P
B
To prove : seg AP seg BP
Construction: Draw seg OA and seg 08.
Proof: In AOPA and AOPB
ZOPA
seg OP chord AB
Ses OP-Seg OP
(common side)
Hypotenuse OB
Hypotenuse
ADA​

Answer 1

Answer:

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.