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Prove

Posted March 20th, 2010 by alpha
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Given that ABC is a triangle such that AB=AC.If D is the mid point of BC , E is the foot of perpendicular from D to AC ,F is the mid point of DE.Then prove that F is perpendicular to BE.

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answer

On March 27th, 2010 Structure says:

A(0,2c)
B(-2b,0)
C(2b,0)
D(0,0)
Line AC

$$\frac{x}{2b}+\frac{y}{2c}-1=0$$

Slope

$$m=-\frac{c}{b}$$

Let

$$a^2=b^2+c^2$$

Line DE has equation

$$y=\frac{b}{c}x$$

Point E

$$E=(\frac{2bc^2}{a^2},\frac{2b^2c}{a^2})$$

Point F

$$F=(\frac{bc^2}{a^2},\frac{b^2c}{a^2})$$

Slope of AF

$$m_{AF}=\frac{\frac{b^2c}{a^2}-2c}{\frac{bc^2}{a^2}}=\frac{b^2-2a^2}{bc}=-\frac{a^2+c^2}{bc}$$

Slope of BE

$$m_{BF}=\frac{\frac{2b^2c}{a^2}}{\frac{bc^2}{a^2}+2b}=\frac{bc}{a^2+c^2}$$
$$m_{AF}m_{BF}=-1$$

Lines are orthogonal.

Please Reply.PLEASEEEEEEEEEEEEEEEEEEEEEE.........

On March 26th, 2010 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

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