http://s28.postimg.org/h7revidfx/Pro_1195_1.png Let u= EA=GK And v= GE=CJ Draw JM //BC ( M on AB ,see sketch) Observe that triangles GKE, JKF, JHC and HGB are similar And triangle JGK congruent to GJH ( per hypotenuses- legs congruence) So MG=HJ=GK= u and BM=CJ= v And GA=u+v=GB => G is the midpoint of AB => GHB is 30-60-90 triangle So BH= 5.sqrt(3) and CH= 10- 5.sqrt(3) and u= JH=2.CH= 20-10.sqrt(3) X= DJ=AG+ GM= 5+ u= 25-10.sqrt(3)
http://s28.postimg.org/h7revidfx/Pro_1195_1.png
ReplyDeleteLet u= EA=GK
And v= GE=CJ
Draw JM //BC ( M on AB ,see sketch)
Observe that triangles GKE, JKF, JHC and HGB are similar
And triangle JGK congruent to GJH ( per hypotenuses- legs congruence)
So MG=HJ=GK= u and BM=CJ= v
And GA=u+v=GB => G is the midpoint of AB => GHB is 30-60-90 triangle
So BH= 5.sqrt(3) and CH= 10- 5.sqrt(3) and u= JH=2.CH= 20-10.sqrt(3)
X= DJ=AG+ GM= 5+ u= 25-10.sqrt(3)
From ∆JGK ≡ ∆GJH => MG=GK ?
Deleteplease explain
Draw GM // AD => GMJH cyclic, From GJ diameter => GMJH kite
ReplyDelete=> G midpoint
GB=5, BH=5√3, HC=10-5√3, FD=20-10√3, x=25-10√3