Question

In the adjoining figure, OD is perpendicular to
the chord AB of a circle with centre O. If BC is a
diameter, show that ACIDO and AC = 2XOD.​

Answer 1

Answer:

your proof is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: prove : \\ 1. \: AC \: || \: DO \\ 2.AC = 2 \times DO \\ \\ so \: here \\ since \: OD \: ⊥ \: AB \: \\ ∠BDO = 90 \: \: \: \: (1) \\ \\ moreover \: since \: here \\ BC \: is \: a \: diameter \\ we \: know \: that \\ diameter \: always \: subtends \: a \\ \: righ t\: angle \: at \: any \: point \: on \: the \: circle \\ therefore \: then \\ ∠BAC = 90 \: \: \: \: \: \: (2) \\ \\ so \: from \: (1) \: and \: (2) \\ we \: conclude \: that \\ ∠BAC = ∠BDO \\ \\ so \: by \: test \: for \: parallel \: lines \\ we \: can \: say \: that \\ AC \: || \: DO \: \: \: \: \: (3)$

$now \: considering \: \\ △BAC \: and \: △BDO \\ ∠CBA = ∠DBO \: \: \: \: \: \: (common \: angle \: ) \\ ∠BAC = ∠BDO \: \: \: \: \: \: \: \: (from \: (3)) \\ henceforth \\ △BAC \: is \: similar \: to \: △BDO \\ by \: AA \: test \: of \: similarity \\ \\ so \: now \\ by \: definition \: of \: \: c.a.s.t \\ we \: obtain \\ \\ \frac{OB}{BC} = \frac{OD}{AC} \\ \\ since \: BC \: is \: a \: diameter \\ and \: OB \: one \: of \: the \: radius \\ we \: have \\ d = 2r \\ BC = 2 \times OB \: \: \: \: \: (4) \\ \\ now \: then \\ \\ \frac{OB}{2.OB} = \frac{OD}{AC} \: \: \: \: \: \: from \: (4)\\ \\ \frac{OD}{AC} = \frac{1}{2} \\ \\ AC = 2.OD \\ thus \: proved$