See complete Problem 36

Right triangle, altitude, incircles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 36

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## Monday, May 19, 2008

###
Elearn Geometry Problem 36

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Online Geometry theorems, problems, solutions, and related topics.

See complete Problem 36

Right triangle, altitude, incircles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

We've BD = ca/b, AD = c^2/b and therefore,

ReplyDeleter1=(ac/b+c^2/b)/(c+ca/b+c^2/b)= ac^2/(b(a+b+c))

Likewise, r2 = ca^2/(b(a+b+c)) and r=ac/(a+b+c)

Hence, LHS = ac^3/(b(a+b+c))+ca^3/(b(a+b+c))

= ac(c^2+a^2)/(b(a+b+c))

= acb/(a+b+c) = r*b since r=ac/(a+b+c)

Thus, r1*c + r2*a = r*b or r1*AB+r2*BC=r*AC

QED

Ajit: ajitathle@gmail.com

As ADB, BDC, and ABC are similar, AB/r1=BC/r2=AC/r=k, so AB=kr1, BC=kr2, AC=kr.

ReplyDeleteFrom AB^2+BC^2=AC^2 we get AB.kr1+BC.kr2=AC.kr and finaly r1.AB+r2.BC=r.AC.

Our proof is based on the simple fact that the in-radius of a triangle XYZ right angled at Z is given by (XZ+YZ-XY)/2.

ReplyDeleteAccordingly,

r1 = (AD + DB - AB)/2 and r = (AB + BC - AC)/2

So r1/r = (AD + DB - AB)/(AB + BC - AC)

= (AB/AC).[AD/AB + DB/AB - 1]/[AB/AC + BC/AC - 1]

= cos A.(cos A + sin A - 1}/(cos A + sin A - 1}

= cos A

Similarly r2/r = cos C

Hence

r1.AB + r2.BC

= r1.c + r2.a

=(r cos A)c + (r cos C)a

= r(a cos C + c cos A)= r.b

= r.AC

By similar triangles ADB and ABC we have

ReplyDeleter1/AD=r/AB

(1.) r1AB=rAD

By similar triangles BDC and ABC we have

r2/DC=r/BC

(2.)r2BC=rDC

Adding equations 1 and 2 and using AD+DC=AC

r1AB+r2BC=r(AD+DC)

r1AB+r2BC=rAC