Triangle and basic properties

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Triangle and basic properties

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Megus

#1 h Nov 11, 2005, 12:38 PM • 1 Y

Y by Adventure10

The points $D, E$ on the side $AB$ of the triangle $ABC$ are such that $\frac{AD}{DB}\frac{AE}{EB} = \left(\frac{AC}{CB}\right)^2$ . Show that $\angle ACD = \angle BCE$ .

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#2 h Nov 11, 2005, 1:23 PM • 1 Y

Y by Adventure10

we have:

$\frac{AE}{EB} = \frac{AC. sin(ACE)}{BC . sin(DCB)}$

and

$\frac{AD}{DB} = \frac{AC. sin(DCB)}{BC . sin(ACE)}$

multipliyng the two results, we have $(\frac{sin(ACE)}{sin(DCB)})^2 = 1$ .

the rest is easy.

E.L

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#3 h Nov 11, 2005, 7:00 PM • 2 Y

Y by Adventure10, Mango247

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=59078 .

darij

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losnak

#4 h Nov 21, 2005, 7:40 PM • 1 Y

Y by Adventure10

<ABC=b <ACB= c <CAB=a <ACD=t <BCE=p <CDE=a+t
<DCE=c-t-p <CED=b+p
we want:p=t

so we have:
(AD/AC)*(AE/AC)=(BD/BC)*(BE/BC) <=>
(sint/sin(a+t))*(sin(c-p)/sin(b+p))=(sin(c-t)/sin(a+t))*(sinp/sin(b+p)) <=>
2sint*sin(c-p)=2sin(c-t)*sinp <=>
cos(c-t-p)-cos(c-p+t)=cos(c-t-p)-cos(c-t+p) <=>
cos(c-p+t)=cos(c-t+p) <=>
c-p+t=c-t+p because they are less than 180º
so 2p=2t <=> p=t

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#5 h Jul 15, 2006, 6:38 PM • 2 Y

Y by Adventure10, Mango247

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