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