Question

in the figure BC is a tangent to the circle with the centre of O. OE bisects AP. prove that ∆AEO~∆ABC.
guys please answer board exam is day after tomorrow

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

we have given a circle with center o and  oe bisect ap

to prove :- ΔAEO ≈  Δ ABC

PROOF  :- WE know a line from center to the chord to its mid point bisect the chord perpendicular to it

therefore  oe ⊥ap

⇒ ∠aeo = 90

and we also know that tangent is ⊥ to radius

thus, ∠abc = 90

now,

in Δ aeo and Δabc

      ∠aeo = ∠abc [ (each 90 ) ( proved above) ]

      ∠oae = ∠ cab ( common )

now,

ΔAEO ≈  Δ ABC { AA SIMILARITY}

                                                   hence prooved