Question Number 14863 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

Commented byadelson last updated on 05/Jun/17

what′s W?

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

ABCD,square.  1)show that:  DE^2 +EB^2 =AE^2 +EC^2  .  2)if :AE=EB=AB⇒∡CED=?

Commented byRasheedSoomro last updated on 05/Jun/17

AE=EB=AB⇒CE=DE  The diagram is not exactly according to  data.

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

mr Rasheed! in part #2, we assume   that: if AE=EB=AB.  for part #2 ,diagram not in scale.  figure of mr Ajfour is correct.

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

if :t=z=a⇒ h_2 =a((√3)/2),h_4 =a−a((√3)/2)  tg∡CDE=(h_4 /(a/2))=((a−a((√3)/2))/(a/2))=2−(√(3.))  ⇒∡CDE=∡EDC=15^•   ⇒∡DEC=180−2×15=150^•  .■

Answered by ajfour last updated on 05/Jun/17

DE^2 +EB^2 =x^2 +y^2 +(a−x)^2 +(a−y)^2   AE^2 +EC^2 =(a−x)^2 +y^2 +x^2 +(a−y)^2   therefore equal.    When △AEB is equilateral,    a−y=y    or    y=(a/2)    a−x= acos (π/6) = ((a(√3))/2)    ⇒   x =a(1−((√3)/2)) = a(((2−(√3))/2))    ∠CED= 2tan^(−1) (((a−y)/x))                   =2tan^(−1) ((y/x))                   =2tan^(−1) [(((a/2))/(a(2−(√3))/2)))]                   =2tan^(−1) (2+(√3))   or ∠CED=π− tan^(−1) ((1/(√3)))                         =π−(π/6)=((5π)/6) .

Commented byajfour last updated on 05/Jun/17

Commented byRasheedSoomro last updated on 05/Jun/17

You′re quicker!

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

thank you mr Ajfour!your answer is  quick and nice and correct and smart!  god bless you my friend.

Commented byajfour last updated on 05/Jun/17

Tanks of thanks Sir.