Let the intersection of $AD,BE$ be $O$. Let $OE=e,OD=D$. Then $OA=2d,OB=2e$. By the Pythagorean Theorem, $e^2+4d^2=AE^2=16,d^2+4e^2=DB^2=9$, adding which yields $d^2+e^2=5$. Thus, $AB=\sqrt{4d^2+4e^2}=\sqrt{4(d^2+e^2)}=\sqrt{4(5)}=2\sqrt{5}$.

BigSams wrote: Let the intersection of $AD,BE$ be $O$. Let $OE=e,OD=D$. Then $OA=2d,OB=2e$. By the Pythagorean Theorem, $e^2+4d^2=AE^2=16,d^2+4e^2=DB^2=9$, adding which yields $d^2+e^2=5$. Thus, $AB=\sqrt{4d^2+4e^2}=\sqrt{4(d^2+e^2)}=\sqrt{4(5)}=2\sqrt{5}$. Could you tell me why, $d^2+e^2=5$

@above Because the point where two medians intersect has to be the centroid and the centroid divides a median in the ratio 2:1

I am going to BARY this problem. Use Barycentrics on $\triangle ABC$ and set $A=(1,0,0),B=(0,1,0),C=(0,0,1),D=(0,\tfrac12,\tfrac12),E=(\tfrac12,0,\tfrac12)$. We have $\overrightarrow{AD}=(-1,\tfrac12,\tfrac12)$ and $\overrightarrow{BE}=(\tfrac12,-1,\tfrac12)$. By EFFT, $\overrightarrow{AD}\perp\overrightarrow{BE} \iff 6^2(\tfrac14-\tfrac12)+8^2(-\tfrac12+\tfrac14)+c^2(1+\tfrac14)=0$. Solving for $c$, our answer is $2\sqrt5$. $\blacksquare$

Let $O$ be the intersection of $AD,BE$ Let $OE=a$ and $OD=b$ $\Rightarrow OB=2a, OA=2b$ Using pythagorean theorem, we get: $a^2+(2b)^2=AE^2 \Rightarrow a^2+4b^2=4^2=16$ $b^2+(2a)^2=BD^2 \Rightarrow b^2+4a^2=3^2=9$ Adding them together, we get: $5a^2+5b^2=25 \Rightarrow a^2+b^2=5$ Finally, we get: $AB^2=(2a)^2+(2b)^2=4a^2+4b^2=4(a^2+b^2)=4 \cdot 5=20 \Rightarrow AB= \sqrt{20} =2 \sqrt{5}$ $\blacksquare$