On the isosceles triangle ABC (AB = BC), on the side AB, we took point D and F (point D is closer to B), and on the side BC

On the isosceles triangle ABC (AB = BC), on the side AB, we took point D and F (point D is closer to B), and on the side BC, point E so that the segments BD = DE = EF = FC = CA Find the angles of the triangle ABC.

After constructing points F, D, E and segments BD = DE = EF = FC = CA, we get a whole set of isosceles triangles: BDE, DEF, EFC and FCA.

We will assume that the angle B is equal to x.

<A = <C = (180 ° – <B) / 2 = 90 ° – x / 2;
<B = <BED = x;
<BDE = 180 ° – <B – <BED = 180 ° – 2x;
<FDE = 180 ° – <BDE = 180 ° – (180 – 2x) = 2x;
<DFE = <FDE = 2x;
<DEF = 180 ° – <FDE – <DFE = 180 ° – 2x – 2x = 180 ° – 4x;
<FEC = 180 ° – <BED – <DEF = 180 ° – x – (180 – 4x) = 3x;
<FCE = <FEC = 3x;
<AFC = <A = 90 ° – x / 2;
<ACF = 180 ° – <A – <AFC = 180 ° – (90 ° – x / 2) – (90 ° – x / 2) = x;
<C = <FEC + <ACF = 3x + x = 4x;
<A = <C;
90 ° – x / 2 = 4x;
180 ° – x = 8x;
9x = 180 °;
x = 20 °;

Answer: <B = 20 °, <A = <B = (180 ° – 20 °) / 2 = 80 °.