Answer:
21cm; 28cm; 14cm
Step-by-step explanation:
There is no info in the problem/s text which one of the triangle's side is 21 cm. That is why we have to try all possible variants.
Let the triangle is ABC . Let the AK is the angle A bisector and BM is median.
Let O is AK and BM cross point.
Have a look to triangle ABM. AO is the bisector and AOB=AOM=90 degrees (means that AO is as bisector as altitude)
=> triangle ABM is isosceles => AB=AM (1)
1. Let AC=21 So AM=21/2=10.5 cm
So AB=10.5 cm as well. So BC= P-AB-AC=63-21-10.5=31.5 cm
Such triangle doesn' t exist ( is impossible), because the triangle's inequality doesn't fulfill. AB+AC>BC ( We have 21+10.5=31.5 => AB+AC=BC)
2. Let AB=21 So AM=21 and AC=42 .So BC= P-AB-AC=63-21-42=0 cm- such triangle doesn't exist.
3. Finally let BC=21 cm. So AB+AC= 63-21=42 cm
We know (1) that AB=AM so AC=2*AB. So AB+AC=AB+2*AB=3*AB
=>3*AB=42=> AB=14 cm => AC=2*14=28 cm.
Let check if this triangle exists ( if the triangle's inequality fulfills).
BC+AB>AC 21+14>28 - correct=> the triangle with the sides' length 21cm,14 cm, 28cm exists.
This variant is the only possible solution of the given problem.
Just square the top and bottom of the fraction. 
and 
so we have 12/16 and then we reduce to 3/4