Li Shanlan's combinatorial identities I am struggling to prove the following combinatorial identities:
$$(1)\quad\sum_{r=0}^m \binom{m}{r}\binom{n}{r}\binom{p+r}{m+n} = \binom{p}{m}\binom{p}{n},\quad \forall n\in\mathbb N,p\ge m,n$$
$$(2)\quad\sum_{r=0}^m \binom{m}{r}\binom{n}{r}\binom{p+m+n-r}{m+n} = \binom{p+m}{m}\binom{p+n}{n},\quad \forall p\in\mathbb N$$
The book I found them in says they were discovered and proved by Chinese mathematician Li Shanlan in the 19th century but I have failed to find any of his works translated into English in the Internet.
I am looking for an either combinatorial or algebraic solution.

Very striking identities. Hard to guess how Li Shanlan discovered them...
Concerning (1), multiply by $x^n y^m z^p$ and add up. 
In the rhs we have
$$
\sum_{p\ge 0}z^p\sum_{n=0}^p {p \choose n} x^n \sum_{m=0}^p {p \choose m} y^m  = \sum_{p\ge 0}(1+x)^p(1+y)^p z^p = \frac{1}{1-z(1+x)(1+y)}.
$$
The lhs is tougher. Denoting $t = \frac{z}{1-z}$,
$$
\sum_{r\ge 0}\sum_{n\ge r}{n\choose r}x^n \sum_{m\ge r}{m\choose r}y^m \sum_{p\ge m+n-r} {p+r\choose m+n}z^p \\
= \sum_{r\ge 0}\sum_{n\ge r}{n\choose r}x^n \sum_{m\ge r}{m\choose r}y^m\frac{z^{m+n-r}}{(1-z)^{m+n+1}} = \sum_{r\ge 0}\frac1{z^r(1-z)}\sum_{n\ge r}{n\choose r}(tx)^n \sum_{m\ge r}{m\choose r}(ty)^m\\
= \sum_{r\ge 0}\frac1{z^r(1-z)}\frac{(tx)^r}{(1-tx)^{r+1}}\frac{(ty)^r}{(1-ty)^{r+1}} = \frac{1}{(1-z)(1-tx)(1-ty)}\frac{1}{1-\frac{1}{z}\frac{tx}{1-tx}\frac{ty}{1-ty}}\\
= \frac{1}{(1-tx)(1-ty)\left(1-z-\frac{txy}{(1-tx)(1-ty)}\right)} = \frac{1}{(1-z)(1-tx)(1-ty) - txy}\\
= \frac{1}{(1-z-xz)(1-ty) - txy} = \frac{1}{1-xz -yz - txy + txyz} \\
= \frac{1}{1-z-xz -yz - xyz} = \frac{1}{1-z(1+x)(1+y)}.
$$
