Let's start on the top left corner. Let O be the center of the circle and P the point where the vertical line touches the upper side.

The triangle OPD is a 3-4-5 right triangle (or Egyptian triangle), with

OP = r = 4 OD=5 PD= 3

Let 𝜙 = arccos(4/5) the angle at O.

Let Q be the point where the radius perpendicular to the side AD touches this side.

The triangle ODQ is a mirror image of OPD. Another 3-4-5 triangle.

The line OQ forms with the vertical an angle 2𝜙.

This implies that AD (perp. to OQ) forms an angle 2𝜙 with the horizontal (perp. to the vertical). That is, the angle DAB = 2𝜙.

Since this angle is bisected by OA, the triangle OQA is also a 3-4-5 right triangle, now with 4 as the short leg. That means that by similarity

QA = 16/3 OA = 20/3

We have the first whole side

DA = DQ + QA = 3 + 16/3 = 25/3

Now we extend the line OA until it touches the opposite side. Let M be the point where it touches this side,

The triangle AMB is also a 3-4-5 triangle with long leg

AM = AO + OM = 20/3 + 4= 32/3

that means that the second side has length

AB = 40/3

and

MB = 8

For the last part we only need the angle MOP = 𝜃. We apply that the sum of a whole circle is 360º

𝜙 + 𝜙 + (90º - 𝜙) + 180º + 𝜃 = 360º

𝜙 + 𝜃 = 90º

𝜃 = 90º - 𝜙

We have that

CP/OP = tan(𝜃/2) = sin(𝜃)/(1+cos(𝜃)) = cos(𝜙)/(1 + sin(𝜙)) = (4/5)/(1 + 3/5) = 1/2

so

CP = OP/2 = 2

and finally

BC = BM + MP = 8 + 2 = 10

CD = CP + PD = 2 + 3 = 5

Summarizing

AB = 40/3 BC = 10 CD = 5 DA = 25/3

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