Right Triangle Altitude 9597Ee

1. **Problem statement:** We have a right triangle $\triangle ABC$ right-angled at $A$. $D$ is the midpoint of the altitude $AH$ from $A$ to $BC$. The perpendicular bisector of $AB$ intersects $CD$ at $E$, and the perpendicular bisector of $AC$ intersects $BD$ at $F$. We need to prove: a) $$\frac{2}{AH} = \frac{1}{BE} + \frac{1}{CF}$$ b) $$BC^2 = 4 \cdot BE \cdot CF$$ 2. **Key concepts and formulas:** - In a right triangle, the altitude to the hypotenuse relates the segments of the hypotenuse. - The midpoint $D$ of $AH$ means $AD = DH = \frac{AH}{2}$. - The perpendicular bisector of a segment is the locus of points equidistant from the segment's endpoints. - We will use coordinate geometry or vector methods to express lengths and prove the relations. 3. **Set up coordinate system:** Place $A$ at the origin $(0,0)$. Let $AB$ lie along the $x$-axis, so $B = (b,0)$. Let $AC$ lie along the $y$-axis, so $C = (0,c)$. Since $\triangle ABC$ is right-angled at $A$, this is consistent. 4. **Coordinates of points:** - $H$ is the foot of the altitude from $A$ to $BC$. The line $BC$ passes through $B(b,0)$ and $C(0,c)$. Its equation is: $$y = -\frac{c}{b}x + c$$ The altitude from $A(0,0)$ is perpendicular to $BC$, so its slope is the negative reciprocal of $-\frac{c}{b}$, which is $\frac{b}{c}$. Equation of altitude $AH$: $$y = \frac{b}{c} x$$ Find $H$ as intersection of $y = \frac{b}{c} x$ and $y = -\frac{c}{b} x + c$: $$\frac{b}{c} x = -\frac{c}{b} x + c$$ Multiply both sides by $bc$: $$b^2 x = -c^2 x + b c^2$$ Bring terms to one side: $$b^2 x + c^2 x = b c^2$$ $$x (b^2 + c^2) = b c^2$$ $$x = \frac{b c^2}{b^2 + c^2}$$ Then $$y = \frac{b}{c} x = \frac{b}{c} \cdot \frac{b c^2}{b^2 + c^2} = \frac{b^2 c}{b^2 + c^2}$$ So $$H = \left( \frac{b c^2}{b^2 + c^2}, \frac{b^2 c}{b^2 + c^2} \right)$$ 5. **Point $D$ is midpoint of $AH$:** Coordinates of $A$ are $(0,0)$, so midpoint $D$ is: $$D = \left( \frac{0 + \frac{b c^2}{b^2 + c^2}}{2}, \frac{0 + \frac{b^2 c}{b^2 + c^2}}{2} \right) = \left( \frac{b c^2}{2(b^2 + c^2)}, \frac{b^2 c}{2(b^2 + c^2)} \right)$$ 6. **Perpendicular bisector of $AB$:** Segment $AB$ has endpoints $A(0,0)$ and $B(b,0)$. Midpoint of $AB$ is $M_{AB} = \left( \frac{b}{2}, 0 \right)$. Since $AB$ is horizontal, its perpendicular bisector is vertical line: $$x = \frac{b}{2}$$ 7. **Line $CD$:** Coordinates: $C = (0,c)$ $D = \left( \frac{b c^2}{2(b^2 + c^2)}, \frac{b^2 c}{2(b^2 + c^2)} \right)$ Slope of $CD$: $$m_{CD} = \frac{\frac{b^2 c}{2(b^2 + c^2)} - c}{\frac{b c^2}{2(b^2 + c^2)} - 0} = \frac{\frac{b^2 c - 2 c (b^2 + c^2)}{2(b^2 + c^2)}}{\frac{b c^2}{2(b^2 + c^2)}} = \frac{b^2 c - 2 c b^2 - 2 c^3}{b c^2} = \frac{-b^2 c - 2 c^3}{b c^2} = -\frac{b^2 + 2 c^2}{b c}$$ Equation of $CD$: $$y - c = m_{CD} (x - 0) \Rightarrow y = c + m_{CD} x$$ 8. **Find $E$ as intersection of perpendicular bisector of $AB$ and $CD$:** Perpendicular bisector of $AB$ is $x = \frac{b}{2}$. Substitute into $CD$: $$y = c + m_{CD} \cdot \frac{b}{2} = c - \frac{b^2 + 2 c^2}{b c} \cdot \frac{b}{2} = c - \frac{b^2 + 2 c^2}{2 c} = \frac{2 c^2 - b^2 - 2 c^2}{2 c} = -\frac{b^2}{2 c}$$ So $$E = \left( \frac{b}{2}, -\frac{b^2}{2 c} \right)$$ 9. **Perpendicular bisector of $AC$:** Segment $AC$ endpoints: $A(0,0)$ and $C(0,c)$. Midpoint $M_{AC} = (0, \frac{c}{2})$. Since $AC$ is vertical, its perpendicular bisector is horizontal line: $$y = \frac{c}{2}$$ 10. **Line $BD$:** Coordinates: $B = (b,0)$ $D = \left( \frac{b c^2}{2(b^2 + c^2)}, \frac{b^2 c}{2(b^2 + c^2)} \right)$ Slope of $BD$: $$m_{BD} = \frac{\frac{b^2 c}{2(b^2 + c^2)} - 0}{\frac{b c^2}{2(b^2 + c^2)} - b} = \frac{\frac{b^2 c}{2(b^2 + c^2)}}{\frac{b c^2 - 2 b (b^2 + c^2)}{2(b^2 + c^2)}} = \frac{b^2 c}{b c^2 - 2 b^3 - 2 b c^2} = \frac{b^2 c}{b c^2 - 2 b^3 - 2 b c^2}$$ Simplify denominator: $$b c^2 - 2 b^3 - 2 b c^2 = b c^2 - 2 b c^2 - 2 b^3 = - b c^2 - 2 b^3 = -b (c^2 + 2 b^2)$$ So $$m_{BD} = \frac{b^2 c}{-b (c^2 + 2 b^2)} = -\frac{b c}{c^2 + 2 b^2}$$ Equation of $BD$: $$y - 0 = m_{BD} (x - b) \Rightarrow y = -\frac{b c}{c^2 + 2 b^2} (x - b)$$ 11. **Find $F$ as intersection of perpendicular bisector of $AC$ and $BD$:** Perpendicular bisector of $AC$ is $y = \frac{c}{2}$. Substitute into $BD$: $$\frac{c}{2} = -\frac{b c}{c^2 + 2 b^2} (x - b)$$ Multiply both sides by $c^2 + 2 b^2$: $$\frac{c}{2} (c^2 + 2 b^2) = -b c (x - b)$$ Divide both sides by $-b c$: $$-\frac{c (c^2 + 2 b^2)}{2 b c} = x - b \Rightarrow -\frac{c^2 + 2 b^2}{2 b} = x - b$$ So $$x = b - \frac{c^2 + 2 b^2}{2 b} = \frac{2 b^2 - c^2 - 2 b^2}{2 b} = -\frac{c^2}{2 b}$$ Therefore $$F = \left( -\frac{c^2}{2 b}, \frac{c}{2} \right)$$ 12. **Calculate lengths $BE$ and $CF$:** - $B = (b,0)$, $E = \left( \frac{b}{2}, -\frac{b^2}{2 c} \right)$ $$BE = \sqrt{\left(b - \frac{b}{2}\right)^2 + \left(0 + \frac{b^2}{2 c}\right)^2} = \sqrt{\left(\frac{b}{2}\right)^2 + \left(\frac{b^2}{2 c}\right)^2} = \sqrt{\frac{b^2}{4} + \frac{b^4}{4 c^2}} = \frac{b}{2} \sqrt{1 + \frac{b^2}{c^2}} = \frac{b}{2} \cdot \frac{\sqrt{c^2 + b^2}}{c}$$ - $C = (0,c)$, $F = \left( -\frac{c^2}{2 b}, \frac{c}{2} \right)$ $$CF = \sqrt{\left(0 + \frac{c^2}{2 b}\right)^2 + \left(c - \frac{c}{2}\right)^2} = \sqrt{\left(\frac{c^2}{2 b}\right)^2 + \left(\frac{c}{2}\right)^2} = \sqrt{\frac{c^4}{4 b^2} + \frac{c^2}{4}} = \frac{c}{2} \sqrt{\frac{c^2}{b^2} + 1} = \frac{c}{2} \cdot \frac{\sqrt{b^2 + c^2}}{b}$$ 13. **Calculate $AH$:** $A = (0,0)$, $H = \left( \frac{b c^2}{b^2 + c^2}, \frac{b^2 c}{b^2 + c^2} \right)$ $$AH = \sqrt{\left(\frac{b c^2}{b^2 + c^2}\right)^2 + \left(\frac{b^2 c}{b^2 + c^2}\right)^2} = \frac{1}{b^2 + c^2} \sqrt{b^2 c^4 + b^4 c^2} = \frac{b c}{b^2 + c^2} \sqrt{c^2 + b^2} = \frac{b c \sqrt{b^2 + c^2}}{b^2 + c^2}$$ 14. **Prove part (a):** Calculate left side: $$\frac{2}{AH} = \frac{2}{\frac{b c \sqrt{b^2 + c^2}}{b^2 + c^2}} = \frac{2 (b^2 + c^2)}{b c \sqrt{b^2 + c^2}} = \frac{2 \sqrt{b^2 + c^2} (b^2 + c^2)}{b c (b^2 + c^2)} = \frac{2 \sqrt{b^2 + c^2}}{b c}$$ Calculate right side: $$\frac{1}{BE} + \frac{1}{CF} = \frac{1}{\frac{b}{2} \cdot \frac{\sqrt{b^2 + c^2}}{c}} + \frac{1}{\frac{c}{2} \cdot \frac{\sqrt{b^2 + c^2}}{b}} = \frac{2 c}{b \sqrt{b^2 + c^2}} + \frac{2 b}{c \sqrt{b^2 + c^2}} = \frac{2 c^2 + 2 b^2}{b c \sqrt{b^2 + c^2}} = \frac{2 (b^2 + c^2)}{b c \sqrt{b^2 + c^2}}$$ This matches the left side, so part (a) is proven. 15. **Prove part (b):** Calculate $4 \cdot BE \cdot CF$: $$4 \cdot BE \cdot CF = 4 \cdot \frac{b}{2} \cdot \frac{\sqrt{b^2 + c^2}}{c} \cdot \frac{c}{2} \cdot \frac{\sqrt{b^2 + c^2}}{b} = 4 \cdot \frac{b}{2} \cdot \frac{c}{2} \cdot \frac{b^2 + c^2}{b c} = 4 \cdot \frac{b c}{4} \cdot \frac{b^2 + c^2}{b c} = b^2 + c^2$$ Since $BC$ is the hypotenuse: $$BC = \sqrt{b^2 + c^2} \Rightarrow BC^2 = b^2 + c^2$$ Therefore, $$BC^2 = 4 \cdot BE \cdot CF$$ which proves part (b). **Final answers:** $$\boxed{\frac{2}{AH} = \frac{1}{BE} + \frac{1}{CF}}$$ $$\boxed{BC^2 = 4 \cdot BE \cdot CF}$$