Convolving with Matched Filter is same as cross-correlation. Suppose say your known signal is $x[n]$ for $0 \le n \le N-1$.

The matched filter is $$ h[n] = x^*[N-n] $$ where $*$ denotes conjugate (considering a generic complex signal). You can drop the conjugate for real signals.

The convolution with matched filtering operation is $$ y[p] = \sum x[l]h[p-l] $$ $$ h[p-l]=x^*[N-(p-l)]=x^*[l-p+N] $$ Therefore $$ y[p]=\sum x[l]x^*[l-p+N]=\sum x[l]x^*[l-\tau] $$ where $\tau = p-N$ is the difference between index $p$ and $N$.

You can already see the matched filter output $y[p]=\sum x[l]x^*[l-\tau]$ is kind of cross correlation operation.

The Matched Filter output at $p=N$ is $$ y[N]=\sum x[l]x^*[l-N+N]=\sum x[l]x^*[l-\tau]\Big|_{\tau=0} $$ So as you can see here, Matched filtering output at $p=N$ which maximizes the SNR is the same as cross correlation output at zero lag, that is $\tau=0$.

Convolving with Matched Filter is same as cross-correlation.

Suppose say your known signal is $x[n]$ for $0 \le n \le N-1$.

The matched filter is $$ h[n] = x^*[N-n] $$ where $*$ denotes conjugate (considering a generic complex signal). You can drop the conjugate for real signals.

The convolution with matched filtering operation is $$ y[p] = \sum x[l]h[p-l] $$ $$ h[p-l]=x^*[N-(p-l)]=x^*[l-p+N] $$ Therefore $$ y[p]=\sum x[l]x^*[l-p+N]=\sum x[l]x^*[l-\tau] $$ where $\tau = p-N$ is the difference between index $p$ and $N$.

You can already see the matched filter output $y[p]=\sum x[l]x^*[l-\tau]$ is kind of cross correlation operation.

The Matched Filter output at $p=N$ is $$ y[N]=\sum x[l]x^*[l-N+N]=\sum x[l]x^*[l-\tau]\Big|_{\tau=0} $$ So as you can see here, Matched filtering output at $p=N$ which maximizes the SNR is the same as cross correlation output at zero lag, that is $\tau=0$.