Question Number 104998 by mathocean1 last updated on 25/Jul/20

Commented bymathocean1 last updated on 25/Jul/20

ABCD and DEFG are squares.  AB=2DE.  Determinate α.

Commented byajfour last updated on 25/Jul/20

many unnecessary constructions  make question confusing..

Answered by 1549442205PVT last updated on 25/Jul/20

Commented by1549442205PVT last updated on 25/Jul/20

  Putting AB=2a⇒DE=a.Denote by I  intersection point of the ray AE and  the circle (O)  From Pithago′s theorem we get  AE=(√(AD^2 +DE^2 )) =a(√5).(1)  From the relation formular in circle we  get EC.ED=EA.EI⇔a^2 =a(√5).EI  ⇒EI=(a/(√5)).  we have ((EI)/(ED))=((EF)/(AE))=(1/(√5)).It follows that  ΔAOE⋍ΔFIE(s.a.s)⇒EIF^(�) =AOE^(�) =135°  On the other hands,AIB^(�) =ADB^(�) =45°  ⇒AIB^(�) +EIF^(�) =45°+135°=180°.This   shows that E,I,B are colinear(q.e.d)

Commented bymathocean1 last updated on 25/Jul/20

thank you sir.

Answered by mr W last updated on 25/Jul/20

Commented bymr W last updated on 25/Jul/20

tan β=(1/3)  tan γ=(1/2)  α=β+γ  tan α=tan (β+γ)=(((1/3)+(1/2))/(1−(1/3)×(1/2)))=1  ⇒α=45°  ∠AIB=α=∠ACB=45°  ⇒I is on the circle.    BD is diameter of circle,  ⇒ΔBDI is right angled triangle,  ⇒DI⊥BI