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

Labels:
congruence,
Pythagoras,
right triangle

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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

This comment has been removed by a blog administrator.

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