BTGmoderatorDC wrote:
In the figure above - Angle ABC = Angle ADC = \(90^o\). Is AD > BC?
(1) Angle CAB = Angle BCA
(2) CD > AB
Source: e-GMAT
$$AD\,\,\mathop > \limits^? \,\,BC$$
$$\left( 2 \right)\,\,\underline {{\rm{IF}}} \,\,AD > BC > 0\,\,\,\,,\,\,\,\,A{D^{\,2}} > B{C^{\,2}}\,\,\,\,\, \Rightarrow \,\,\,\,\, - A{D^{\,2}} < - B{C^{\,2}}\,\,\,\,\left( * \right)$$
$$C{D^{\,2}} = A{C^{\,2}} - A{D^{\,2}}\,\,\mathop < \limits^{\left( * \right)} \,\,\,A{C^{\,2}} - B{C^{\,2}} = A{B^{\,2}}\,\,\,\,\,\,\mathop \Rightarrow \limits^{CD\,,\,AB\,\, > \,\,0} \,\,\,\,CD < AB\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( 2 \right)\,\,{\rm{contradicted}}\,$$
Conclusion: AD > BC is impossible, hence (2) is SUFFICIENT.
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.