Draw the circumcircle. Let $D$ be the point diametrically opposite $A$.

Extend $BI$ and $BG$ to $AC$ and $AI$ and $AG$ to $BD$. Check out the similar triangles.

The centroid divides a cevian in a 2:1 ratio. Abuse this fact to your advantage.

Angle bisector theorem on $\angle B$.

WLOG let $A$ be the right angle, and $IG \parallel BC$. Let $O$ be the circumcentre of $\triangle ABC$, and let $D$ be the point diametrically opposite $A$. Extend $BI$ to $AC$ at $P$ and $BG$ to $AC$ at $M$. (This means $M$ is the midpoint of $AC$.) Additionally, extend $AI$ to $BD$ and $N$. [asy][asy] unitsize(2); import olympiad; pair A, B, C, D, O, I, G, N, P, M; A=(54,72); B=(0,0); C=(150,0); D=(96,-72); P=(90,45); M=(102,36); N=(72,-54); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw(P--B--M); draw(N--A--D); draw(N--P); draw(M--D); O=circumcenter(A,B,C); I=incenter(A,B,C); G=centroid(A,B,C); dot(I); label("I",I,WNW); dot(G); dot(O); dot(A); dot(B); dot(C); dot(D); dot(N); dot(P); dot(M); label("G",G,ESE); label("A",A,NW); label("B",B,W); label("C",C,E); label("D",D,SE); label("N",N,SSW); label("P",P,ENE); label("M",M,ENE); label("O",O,NE); draw(I--G, green); draw(P--M, green); draw(N--D, green); draw(circumcircle(A,B,C)); [/asy][/asy] First, we notice that the green segments are all parallel to each other. This gives us a lot of similar triangles to work with. Additionally, we recognize that $ABDC$ is a rectangle. Observe that $\triangle IBG$ and $\triangle PBM$ are similar because of angles. Because $\dfrac{BG}{BM}=\dfrac{2}{3}$, we have $PM=\dfrac{3}{2}\cdot 10$ = $15$. Thus, $MC=MA=15+AP$. Next, observe that $\triangle IAG$ and $\triangle NAD$ are also similar because of angles. We have $\dfrac{AG}{AD}=\dfrac{\frac{2}{3}\cdot AO}{2\cdot AO}=\dfrac{1}{3}$, so $ND=3\cdot 10 = 30$. Now, because $ND=2PM$, we see that $BN=BD-ND=2AM-2PM=2AP$, and because $\angle BAN = 45^{\circ}$ and $\angle ABN = 90^{\circ}$, $AB=BN=2AP$. Finally, through the angle bisector theorem, \begin{align*} \dfrac{AB}{AP}&=\dfrac{CB}{CP} \\ \Longrightarrow \dfrac{2AP}{AP}&=\dfrac{CB}{AP+30} \\ \Longrightarrow CB&=2AP+60 \\ \end{align*}and through the Pythagorean theorem, \begin{align*} AB^2+AC^2&=BC^2 \\ \Longrightarrow (2AP)^2+(2AP+30)^2&=(2AP+60)^2 \\ \end{align*}Solving gives $AP=45$, so $AB=2AP=90$ and $AC=2AP+30=120$. Thus, the two legs have lengths $\boxed{90\text{ and }120}$. Yay! :DD

Define $ABC$ as the triangle and let $BC,CA,AB$ be $a,b,c$ respectively. $a^2+b^2=c^2$ of course. WLOG $IG$ is parallel to $BC$. This also means $IG$ is perpendicular to $AC$. The altitude of $I$ wrt $BC$ is $r=\frac{a+b-c}2$, and the altitude of $G$ wrt $BC$ is $\frac b3$. Because $IG$ is parallel to $BC$, these altitudes must be equal. $$\frac{a+b-c}2=\frac b3\implies3a+b=3c\implies(3a+b)^2=9(a^2+b^2)\implies3a=4b$$This shows the right triangle is 3-4-5. Let $a,b,c$ be $4x,3x,5x$ respectively. The altitude of $I$ wrt $AC$ is $r=\frac{a+b-c}2=x$, and the altitude of $G$ wrt $AC$ is $\frac a3=\frac{4x}3$. Therefore $IG=\frac{4x}3-x=\frac x3$ and as $IG=10$, we obtain $\frac x3=10\implies x=30$. The lengths of the other two legs of the triangle are $\boxed{BC=120\text{ and }AC=90}$.