swerve wrote:$$\text{In the figure given below, } AB = BC = 2\sqrt{2} \text{ units and } AC = 4 \text{ units.}$$ $$\text{If } BD \text{ bisects the side } AC\text{, find the length of }BD.$$
$$\text{A. } 1 \, \, \, \text{B. } \sqrt{2} \, \, \, \text{C. }2 \, \, \, \text{D. } 2\sqrt{2} \, \, \, \text{E. } 3$$
\[? = BD\]
\[{\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2}\,\,\,\, \Rightarrow \,\,\,\,\Delta ABC\,\,right\,\,,\,\,\,\angle ABC = {90^ \circ }\]
\[\frac{{AC \cdot BD}}{2} = {S_{\,\Delta \,ABC\,}} = \frac{{AB \cdot BC}}{2}\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{4 \cdot ?}}{{}} = \frac{{2\sqrt 2 \cdot 2\sqrt 2 }}{{}}\,\,\,\,\,\, \Rightarrow \,\,\,\,? = 2\]
Obs.: the reason that guarantees BD perpendicular to AC is symmetry or, if you prefer, the fact that AC is a base of an isosceles triangle, therefore
relative to the base, the median (given) coincides with the height.
The above follows the notations and rationale taught in the GMATH method.
