Question

AD is the median of triangle ABC , O is any point on AD. BO and CO produced meet AC and AB in E and F respectively. AD is produced to X such that OD=DX. Prove that AO: AX = AF : AB​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: prove \: that : \\ AO:AX=AF:FB \\ \\ so \: considering \\ △AFO \: and \: △ABX \\ ∠FAO=∠BAX \: \: \: \: \:( common \: angle \: ) \\ ∵ \: FO \: || BX \\ ∠AFO=∠ABX \: \: \: \: (corresponding \: angles) \\ \\ ∴ \: △AFO \: \: is \: similar \: to \: △ABX \\ by \: AA \: \: test \: of \: similarity \\ \\ thus \: then \\ \frac{AO}{AX} = \frac{AF}{FB} \: \: \: \: (c.s.s.t) \\ \\ AO:AX=AF : FB \\ thus \: proved$