Question

help me to do this sum​

Answer 1

Answer:

hey here is your answer

pls mark it as brainliest pls

Step-by-step explanation:

$so \: here \: BX \: and \: CY \: are \: angle \: bisectors \: of \: ∠ABC \: and \: ∠ACB \: respectively \\ so \: hence \: \\ ∠ABX = ∠CBX = \frac{1}{2} ∠ABC \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1) \\ \: similarly \: \: ∠ACY = ∠BCY = \frac{1}{2} ∠ACB \: \: \: \: \: \: \: \: \: (2)$

$so \: here \: AB = AC \\ so \: by \: isosceles \: triangle \: theorem \\ we \: get \\ ∠ABC = ∠ACB \: \: \: \: \: \: \: \: (3) \\ so \: from \: (1) \: and \: (2) \\ we \: get \\ ∠ACY = ∠BCY = ∠ABX= ∠CBX \\ ie \: considering \: ∠ABX = ∠ACY \: \: \: \: \: \: \: (4)$

$now \: here \: as \: AB=AC \\ AB \: and \: AC \: both \: are \: also \: chords \: of \: given \: circle \: \\ so \: we \: know \: that \\ whenever \: chords \: are \: equal \: their \: corresponding \: arcs \: are \: equal \\ hence \: \: then \\ m(arc \: AB)=m(arc \: AC) \: \: \: \: \: \: \: \: (5)$

$so \: now \: here \\ ∠AXB \: and \: ∠AYC \: intercepts \: arc \: AB \: and \: AC \: repectively \: on \: circumference \: of \: circle \\ so \: by \: applying \: inscribed \: angle \: theorem \\ we \: get \\ ∠AXB = \frac{1}{2} .m(arc \: AB) \: \: \: \: \: \: \: \: \: \: \: (6) \\ \\ ∠AYC = \frac{1}{2} .m(arc \: AC) \: \: \: \: \: \: \: \: \: \: (7) \\ \\ so \: from \: (5) \: (6) \: and \: (7) \\ we \: get \\ ∠AYC = ∠AXB \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (8)$

$now \: considering \: \\ △AXB \: and \: △AYC \\ ∠ABX= ∠ ACY \: \: \: \: \: \: \: \: \: \: from \: (4) \\ ∠AXB= ∠AYC \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: from \: (5) \\ thus \: then \: \\ △AXB \: is \: similar \: to \: △AYC \: \: \: \: \: \: \: \: \: \: (AA \: test \: of \: similarity) \\ hence \\ \frac{AB}{ AC } = \frac{BY}{ AX} \: \: \: \: \: \: \: (c.s.s.t) \\ so \: as \: AB = AC \\ we \: get \\ 1 = \frac{BY}{ AX } \\ ie \: BY =AX \\ so \: as \: BY = 4.cm \: \: \: \: \: \: \: \: (given) \\ AX = 4.cm$