Question

Answer this question ​

Answer 1

Answer:

O is the centre and OM⊥AB

Prove that AM=BM

${\huge{\underline{\small{\mathbb{\pink{REFER \ TO \ THE \ ATTACHMENT}}}}}}$

Statement:-

In triangle OMA and OMB,

• OA=OB
• ∠OMA=OMB
• OM=OM

$\therefore \triangle \: oma \cong \triangle \: omb \\ \\ \therefore \: am = bm$

Reason:-

• Radii of same circle both 90°
• Common side by RHS C.S.C.T

$\rule{200pt}{2.5pt}$

Hope it helps you from my side

:)

Answer 2

Answer:

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.