Question 979057
In a right triangle, the altitude to the hypotenuse splits the triangle into two smaller right triangles, and you end up with three similar right triangles
As a consequence, there is a bunch of proportions between the lengths of the sides of those triangles that are useful to solve this problem.
You may even have had to memorize some formulas/theorems based on that.
{{{drawing(400,200,-1,17,-1.5,7.5,
triangle (0,0,4,0,4,6.93),triangle (0,0,16,0,4,6.93),
rectangle(4,0,4.5,0.5),locate(7.85,-0.1,a),
arrow(7.7,-0.5,0,-0.5),arrow(8.3,-0.5,16,-0.5),
locate(1.9,0.7,d),locate(9.9,0.7,e),
locate(2,3.45,b),locate(9.5,3.45,c),locate(4.1,3.5,h),
line(3.75,6.5,4.18,6.25),line(4.43,6.68,4.18,6.25)
)}}} {{{b/a=d/b}}}<--->{{{b^2=ad}}} , and {{{c/a=e/c}}}<--->{{{c^2=ae}}} .
In this problem there is another perpendicular to the hypotenuse, like this
{{{drawing(400,200,-1,17,-1.5,7.5,
triangle (0,0,4,0,4,6.93),triangle (0,0,16,0,4,6.93),
rectangle(4,0,4.5,0.5),triangle (10,0,16,0,10,3.465),
rectangle(10,0,10.5,0.5),locate(7.85,-0.6,a),
arrow(7.7,-1,0,-1),arrow(8.3,-1,16,-1),
locate(1.9,0.7,d),locate(2,3.45,b),
locate(12,2.3,c/2),locate(6.8,5.2,c/2),
locate(4.05,3.5,h),
locate(12.3,0,"e/2"),locate(6.3,0,"e/2"),
line(3.75,6.5,4.18,6.25),line(4.43,6.68,4.18,6.25)
)}}}
That perpendicular is splitting in half the hypotenuse and one leg of one of the smaller triangles.
In the problem  {{{a=10cm+6cm=16cm}}} ,
and it must be that {{{e/2=6cm}}}<-->{{{e=2*6cm=12cm}}} ,
because {{{e/2=10cm}}}-->{{{e=20cm>16cm=a}}} does not make sense.
Since {{{e=12cm}}} , then {{{d=a-e=16cm+12cm=4cm}}}
Using the proportions above,
{{{b^2=ad}}}-->{{{b=sqrt(ad)=sqrt((16cm)(4cm))=sqrt(64cm^2)=highlight(8cm)}}} and
{{{c^2=ae}}}-->{{{c=sqrt(ae)=sqrt((16cm)(12cm))=sqrt(192cm^2)=highlight(sqrt(3)cm)=about13.86cm}}}