If $W(t)$ is a standard Wiener process (i.e., for $t \geq 0, W(t)$ is normally distributed with expected value 0 and variance $t,$ and the increments are stationary and independent), then $$ B(t)=W(t)-\frac{t}{T} W(T) $$ is a Brownian bridge for $t \in[0, \mathrm{T}]$.
How to show that $B$ is a Markov process ? I have 2 ways in mind: some sort of direct proof or showing that $B$ is an Ito process (then it is automatically Markov). I would like to see both.
Edit: this is not correct by @UBM's comment
We have $dB_t = dW_t - \frac{W(T)}{T}dt$, so $B_t$ is an Ito process.
To prove directly that $B_t$ is Markov is essentially the same. Note that $B_t$ is a sum of a Wiener process and a term that depends exclusively on time. The latter term is clearly independent of the filtration, since it is deterministic. The former is known to be independent of the filtration since it has independent increments. It isn't too difficult to convince oneself that in this case this makes $B_t$ independent of the filtration.