= Solution
For <bremsstrahlung>, the <electron> scatters from the <nucleus> by virtual-photon exchange and emits a real <photon>: $e^-Z\to e^-\gamma Z$. The two lowest-order graphs have emission before or after the Coulomb scattering. One is drawn in the figure; interchanging the emission and exchange vertices gives the other. Their sum is needed for the full gauge-consistent <Probability amplitude>[amplitude]. Reversing the charged-lepton arrows gives <Positron> <bremsstrahlung>, with the same underlying electromagnetic mechanism.
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