Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 757 details.
Denote S(XYZ)= area of triangle XYZWe have S(ADB)=S(ADE)+S(DEB)Or ½.b.c.sin(120)=1/2.b.x.sin(60)+1/2.c.x sin(60)Note that sin(60)=sin(120)So b.c=b.x+c.x => x=b.c/(b+c)
Tr.ADE///Tr.CDB since /_CDB=/_ADE both=60 and /_DAE=/_DCB (same sector); hence x/b = c/CD --(1)By Ptolemy or by Problem 256, CD=b+c --(2)Combine (1)&(2) to get, x/b=c/(b+c) or 1/x=1/b+1/c
By similar triangles (ADE ~ BEC)&(BDE ~ AEC),x/b = BE/BC, x/c = AE/ACSince ABC is equiliteral, (BE + AE = AC = BC)Then x/b + x/c = BE/BC + AE / AC = 1The remaining could easily done by rearranging of termsQ.E.D.
By Sine Rule,x/b + x/c = sin DAE / sin AED + sin DBE / sin DEB = sin DCB / sin BEC + sin DCA / sin AEC= BE/AC + EC/AC= (BE + EC) / AC= BC/AC= 1 etc
A pure geometric solution:Let be P on ray AD such that DP=DB. As ADBC is cyclic, x=bc(b+c) QEDVideo solution is coming, Go-Greetings :)
WTF, what happened with my post?!?!!?i will complete this: (...) cylic, triangles ADE and APB are similar, then (...)Go-greetings!