Diagram of Right Triangles
Given: # \frac{\overline{AB}}{\overline{BC}} = \frac{\overline{CD}}{\overline{AC}} = \frac{\overline{AD}}{\overline{DE}} = k #
Required: Find # (\frac{\overline{AE}}{\overline{BC}})^2 #
Analysis: use Pythagorean Theorem #c = \sqrt{a^2 + b^2} #
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Solution: Let, # \overline{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} #
