Question Number 115056 by A8;15: last updated on 23/Sep/20

Answered by 1549442205PVT last updated on 23/Sep/20

 (x/y)=(√2)  Given CF=x,BC=AD=y,CED^(�) =90°  We have DCA^(�) =ECF^(�) (hypothesis)  ⇒Δ_r ADC∽Δ_r FEC⇒((EF)/(CE))=((AD)/(CD))=(y/(CD))(1)  But ((EF)/(CE))=((DF)/(CD))(2)(the property of the  bisector in triangle CDE)  .From (1(2)we get  DF=y=AD⇒ΔADF is isosceles at D  Put DAF^(�) =DFA^(�) =ϕ⇒ { ((DFC^(�) =180−ϕ)),((DCF^(�) =90−ϕ)),((ADF^(�) =180−2ϕ)) :}  ⇒CDF^(�) =90−(180−2ϕ)=2ϕ−90(∗)  For ΔCDF the sine theorem gives us  ((CD)/(sinDFC^(�) ))=((DF)/(sinDCF^(�) ))=((CF)/(sinCDF^(�) ))⇔  ((CD)/(sinϕ))=(y/(cosϕ))=(x/(−cos2ϕ)) (since sin(2ϕ−90  =−sin(90−2ϕ)=−cos2ϕ  and   sin(180−ϕ)=sinϕ,sin(90−ϕ)=cosϕ  ⇒x=((−ycos2ϕ)/(cosϕ))=((y(1−2cos^2 ϕ))/(cosϕ))  ⇒(x/y)=((1−2cos^2 ϕ)/(cosϕ))(3)  In right triangle CEF we get   CE=xsinϕ,note (∗)  BCE^(�) =CDE^(�) =(2ϕ−90)(two angles   have perpendicular coresponding  sides).Hence,in the  right triangle CBE we get  BE=CEsin(2ϕ−90)=−xsinϕcos2ϕ  =BC.tan(2ϕ−90)=−ycot2ϕ  ⇒−xsinϕcos2ϕ=−ycot2ϕ  ⇔(x/y)=((cot2ϕ)/(sinϕcos2ϕ))=(1/(2sin^2 ϕcosϕ))(4)  From(3)(4)we get  ((1−2cos^2 ϕ)/(cosϕ))=(1/(2(1−cos^2 ϕ)cosϕ))=(x/y)  ⇒2(1−2cos^2 ϕ)(1−cos^2 ϕ)=1  Put cos^2 ϕ=t we have  2(1−2t)(1−t)=1⇔4t^2 −6t+1=0  Δ′=5⇒t=((3−(√5))/4)=((6−2(√5))/8)=((((√5)−1)^2 )/8)  ⇒cos^2 ϕ=((((√5)−1)^2 )/((2(√2))^2 ))⇒cosϕ=(((√5)−1)/(2(√2)))  .Replace into (3)we get  (x/y)=((1−2.((3−(√5))/4))/(((√5)−1)/(2(√2))))=(((√(2())(√5)−1))/( ((√5)−1)))=(√2)

Commented byA8;15: last updated on 23/Sep/20


Commented byA8;15: last updated on 23/Sep/20

please help me with the solution