I don't have the course notes, but thought I'd try to help anyway.
The variance V of a distribution is an integral (or sum, same thing) ranging over all the values of the distribution.
You can split this into two integrals: one which sums over the bits below the mean, and another which sums over the bits above the mean. The first integral we call the 'downside semi-variance' D and the second I'm going to call the 'upside semi-variance' U (not sure if this is standard). If you add these two together this is equal to the original variance V, that is D + U = V.
If the downside semi-variance is half the variance, as you've said it is in this question, then 2 * D = V. Hence D = U, so there must be some kind of symmetry around the mean of the distribution as the integral for above the mean is equal to the integral for below the mean.
Any symmetric distribution would achieve D = U (let me know if this isn't obvious), so anything like a Student t or the normal distribution will have the downside semi-variance being half the variance. Boom.
Might be totally off here as I don't have any of the notes, but hopefully I said something right and this helps.
