Inner angles

The magnitude of the internal angle at the main vertex C of the isosceles triangle ABC is 72°. The line p, parallel to the base of this triangle, divides the triangle into a trapezoid and a smaller triangle. How big are the inner angles of the trapezoid?

Correct result:

A =  54 °
B =  54 °
C =  126 °
D =  126 °

Solution:

$$\begin{gathered} \gamma =72^{\circ } \\ \alpha =\beta \\ \alpha +\beta +\gamma =180^{\circ } \\ \alpha +\alpha +\gamma =180^{\circ } \\ \alpha ={\displaystyle \frac{180-\gamma }{2}}={\displaystyle \frac{180-72}{2}}=54^{\circ } \\ \beta =\alpha =54=54^{\circ } \\ A=\alpha =54=54^{\circ } \end{gathered}$$

$B=\beta =54=54^{\circ }$

$$\begin{gathered} C+B=180^{\circ } \\ C=180-B=180-54=126^{\circ } \end{gathered}$$

$$\begin{gathered} A+D=180^{\circ } \\ D=C=126=126^{\circ } \end{gathered}$$

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