Two "triangles" in the right triangle

The problem

This problem is connected with my post to geometry-puzzles newsgroup,
and with my previous problem No. 42.
Solve the problem No. 42,
for the case when the triangle ABC has the right angle ACB, legs a=BC, b=CA, and hypotenuse c=AB, see the Figure: [Graphics:Images/RT-2.jpg]

First, do next six easy construction steps:
1A. Draw the circle with center at A, and with radius b=AC;
1B. draw the circle with center at B , and with radius a=BC ;
1C. draw the circle with center at C, and with radius CD, height to hypotenuse AB;
2A. find point E (inside the triangle ABC!) of intersection of circles in 1B and 1C;
2B. find point F (inside the triangle ABC!) of intersection of circles in 1A and 1C;
2C. the vertex C is point (inside the triangle ABC!) of intersection of circles in 1A and 1B.

For any given right triangle ABC, the triangle with vertices C, E, F is unique (and easy to construct).
Also the "curved-side" triangle with the same vertices C, E, F but with arc sides CJE, EDF, FMC is unique (and easy to construct).

Now assuming that we know side lengths of triangle ABC:
1. Find the area (and perimeter) of the triangle CEF.
2. Find the area (and perimeter) of the "triangle" CJEDFMC.

The hint

N/A.

The answer

See here, but better to try yourself!

The Solution

N/A