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Let ABC be a triangle. Let BE and CF be internal angle bisectors of B and C respectively with E on AC and F on AB. Suppose X is a point on the segment CF such that AX is perpendicular to CF; and Y is a point on the segment BE such that AY perpendicular BE. Prove that XY = $\frac{b+c-a}{2}$, where BC = a , CA = b, AB = c. Please give a proof using coordinate geometry.

My solution: I took a general triangle with two vertices lying on x-axis and one vertex lying on y-axis. Found the equation of angle bisectors of B and C. Found the foot of perpendicular from A on CF and BE as X and Y respectively. Found XY but the answer was not matching . Please help. Again I request, please give a proof using coordinate geometry only.

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  • $\begingroup$ Somebody, please reply $\endgroup$ – user2369284 Nov 30 '13 at 14:12
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The following is a diagram. And $B$ is the origin and assume each point $P$ has coordinate $(p_1,p_2)$.

enter image description here

  • Point $A,B,C$

From the length three edges. We can immediately write $B(0,0)$ and $C(a,0)$. Then solve $\displaystyle A(\frac{a^2-b^2+c^2}{2a},\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{2a})$ from $a_1^2+a_2^2=c^2,~(a_1-a)^2+a_2^2=b^2$.

  • Line $BE,CF$

Noting that $$\displaystyle k_{BE}=\tan\frac B2=\frac{\tan B}{1+\sqrt{1+\tan^2B}}=\frac{k_{AB}}{1+\sqrt{1+k_{AB}^2}}=\frac{a_2}{a_1+\sqrt{a_1^2+a_2^2}}=\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{(a+c)^2-b^2}$$ $$\displaystyle k_{CF}=\frac{-a_2}{(a-a_1)+\sqrt{(a-a_1)^2+(-a_2)^2}}=\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{(a+b)^2-c^2}$$

The equation of $BE$ is $k_{BE}x-y=0$ and $CF$, considering the line passes point $C$, $k_{CF}(x-a)-y=0$

  • Point $X,Y$

Let $X$ be $\displaystyle (x_1,k_{CF}(x_1-a))$, $Y$ be $\displaystyle (y_1,k_{BE}y_1)$ since they are on lines $CF$, $BE$ respectively. According to orthogonal relation, we have

$\displaystyle AX\perp CF\Longleftrightarrow k_{AX}k_{CF}=\frac{a_2-k_{CF}(x_1-a)}{a_1-x_1}k_{CF}=-1\\ \displaystyle AY\perp BE\Longleftrightarrow k_{AY}k_{BE}=\frac{a_2-k_{BE}y_1}{a_1-y_1}k_{BE}=-1$

We obtain $\displaystyle X(\frac{3a^2-2ab-b^2+c^2}{4a},\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{4a}),Y(\frac{a^2-b^2+2ac+c^2}{4a},\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{4a})$ and hence the distance $$d(X,Y)=y_1-x_1=\frac {b+c-a}2$$

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  • $\begingroup$ Why didn't you choose the vertices on the axes? Wouldn't it have been easier to solve ? $\endgroup$ – user2369284 Nov 30 '13 at 14:44
  • $\begingroup$ @user2369284 It could be possible but not have to. You could try that and check everything again. $\endgroup$ – Shuchang Nov 30 '13 at 14:52
  • $\begingroup$ While finding the coordinates of A, we can simply subtract the 2 equations. But I think you have done something else $\endgroup$ – user2369284 Nov 30 '13 at 15:07
  • $\begingroup$ @user2369284 That's right. $\endgroup$ – Shuchang Nov 30 '13 at 15:11
  • $\begingroup$ And what have you done ? $\endgroup$ – user2369284 Nov 30 '13 at 15:11

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