See complete Problem 58
Right Triangle, Pythagoras, Congruence. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 58
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Es simple:
ReplyDeletetraze la altura DH
el angulo BAC es el mismo que el angulo DHC
como AC=DC entonces los triangulos rectangulos ABC y DHC son iguales entonces HC=c,BH=a-c,DH=a
finalmente en el triangulo rectangulo BHD use teorema de pitagoras y se tiene
(a-c)2+a2=x2
The solution is posted on http://www.osinfofrom.us/solution_problem58.html
ReplyDeleteLet angle DCB = z Then we've, z= angle BAC &
ReplyDeletecos(z) = cos(BAC) = c/b
Thus, x^2= a^2 + b^2 - 2a*b*c/b since DC = b
= a^2 + a^2 + c^2 -2a*c since b^2=a^2+c^2
= a^2+(a-c)^2
QED
Ajit: ajitathle@gmail.com
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ReplyDeleteConsider pt.F on BC such that B-F-C and FC=c ..(1)
ReplyDeleteangle BAC=angle FCD...(2)
AC=DC=b...given...(3)
By (1),(2),(3) and SAS test of congruency,
tri ABC is congruent to tri FCD
hence,DF=a...(c.s.c.t) and
angle DFC = angle DFB = 90
Bf = a-c...(by(1))
In tri DFB by pythagoras theorem,
DE^2 + BE^2 = BD^2
a^2 + (a-c)^2 = x^2
Extend AB to to E such thar AE equal a.Then AEC Tr congruent to BCD Tr.SAS.So CE equal x and the result folliws
ReplyDeleteSumith Peiris Sri Lanka