Diagram of Right Triangles
Given: \frac{\overline{AB}}{\overline{BC}} = \frac{\overline{CD}}{\overline{AC}} = \frac{\overline{AD}}{\overline{DE}} = k ¯¯¯¯¯¯AB¯¯¯¯¯¯BC=¯¯¯¯¯¯CD¯¯¯¯¯¯AC=¯¯¯¯¯¯AD¯¯¯¯¯¯DE=k
Required: Find (\frac{\overline{AE}}{\overline{BC}})^2 (¯¯¯¯¯¯AE¯¯¯¯¯¯BC)2
Analysis: use Pythagorean Theorem c = \sqrt{a^2 + b^2} c=√a2+b2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Solution: Let, \overline{BC} = x ¯¯¯¯¯¯BC=x, \because \frac{\overline{AB}}{\overline{BC}} = k,
\overline{AB} = kx , use Pythagorean Theorem to find the value of \overline{AC} :
\overline{AC} = \sqrt{\overline{BC}^2 + \overline{AB}^2 }
= \sqrt{ x^2 + k^2x^2} = \sqrt{(x^2)(1+k^2)} = x\sqrt{1+k^2}
\overline{AC} = x\sqrt{1+k^2}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\because \frac{\overline{CD}}{\overline{AC}} = k, \overline{CD} = \overline{AC} * k = xk\sqrt{1+k^2}
Use Pythagorean Theorem to find the value of \overline{AD} :
\overline{AD} = \sqrt{\overline{CD}^2 + \overline{AC}^2
= \sqrt{(xk\sqrt{1+k^2})^2 + (x\sqrt{1+k^2})^2 }
= \sqrt{x^2k^2(1+k^2) +x^2(1+k^2) }
= \sqrt{x^2 [k^2(1+k^2) + 1(1+k^2)] }
= x\sqrt{(k^2+1)(1+k^2)} , thus
\overline{AD} = x(1+k^2)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\because \frac{\overline{AD}}{\overline{DE}} = k,
\overline{DE} = \frac{\overline{AD}}{k} = \frac{x}{k} * (1+k^2)
Use Pythagorean Theorem to find the value of \overline{AE} :
\overline{AE}^2 = \sqrt{\overline{DE}^2 +\overline{AD}^2 =
= \sqrt{(frac{x}{k} * (1+k^2))^2 + (x(1+k^2))^2
= \sqrt{(x^2/k^2)(1+k^2)^2 + (x^2)(1+k^2)^2
= x\sqrt{(1/k^2 + 1)(1+k^2)^2
= x\sqrt{\frac{1+k^2}{k^2} (1+k^2)^2}
Thus,
\overline{AE} = x\sqrt{\frac{(1+k^2)^3}{k^2}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
(\frac{\overline{AE}}{\overline{BC}})^2
= (\frac{x\sqrt{\frac{(1+k^2)^3}{k^2}}}{x} )^2
= (\sqrt{\frac{(1+k^2)^3}{k^2}})^2
Thus,
(\frac{\overline{AE}}{\overline{BC}})^2 = \frac{(1+k^2)^3}{k^2}
