\(
d+\overline{PT}=b-\overline{OL}\Longrightarrow b-d=\overline{OL}+\overline{PT}\\
\dfrac{\overline{OL}}{\overline{OM}}=\dfrac{a}{\overline{AM}}\Longrightarrow \overline{OL}=a\cdot \dfrac{\overline{OM}}{\overline{AM}}\\
\dfrac{\overline{PT}}{\overline{MP}}=\dfrac{a}{\overline{AM}}\Longrightarrow \overline{PT}=a\cdot \dfrac{\overline{MP}}{\overline{AM}} \)
Por tanto, \( b-d=a\cdot \dfrac{\overline{OM}}{\overline{AM}}+a\cdot \dfrac{\overline{MP}}{\overline{AM}}=a\cdot \dfrac{\overline{OM}+\overline{MP}}{\overline{AM}}=a\cdot \dfrac{2\overline{AM}}{\overline{AM}}=2a \)
ya que \( \overline{MP}=\overline{MD}+\overline{AM}-\overline{OM}=2\overline{AM}-\overline{OM} \)