# Internal angles

The ABCD is an isosceles trapezoid, which holds:

|AB| = 2 |BC| = 2 |CD| = 2 |DA|:

On its side BC is a K point such that |BK| = 2 |KC|, on its side CD is the point L such that |CL| = 2 |LD|, and on its side DA the point M is such that | DM | = 2 |MA|. Determine the internal angles of the KLM triangle.

|AB| = 2 |BC| = 2 |CD| = 2 |DA|:

On its side BC is a K point such that |BK| = 2 |KC|, on its side CD is the point L such that |CL| = 2 |LD|, and on its side DA the point M is such that | DM | = 2 |MA|. Determine the internal angles of the KLM triangle.

**Result****Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):**

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**Math student**

Help. First, look at the inner angles of the ABCD trapezoid.

Solution. It follows from the assumptions that the center line of the AB segment with the vertices C and D divides the ABCD trapezoid into three identical equilateral triangles. Therefore, the magnitude of internal angles in the trapezoid at A and B vertices is equal to 60 °

And at the C and D vertices 120 °. It follows from the specification that the triangles LCK and MDL are the same (according to the sentence above). Therefore, both the KL and LM lines and the designated pairs of angles are the same; The magnitudes of these angles are denoted α and β. The triangle KLM is isosceles and the angles at the base are the same; Their size is denoted by δ and the size of the angle KLM is denoted by γ.

From the sum of the inner angles in the KCL triangle we derive

α + β = 180° − 120° = 60°

The sum of the three marked angles with the vertex L is a straight angle, therefore

γ = 180° − (α + β) = 120°

Finally, we deduce the sum of inner angles in the triangle KLM

δ = (180° − 120°)/2 = 30°

The internal angles of the triangle KLM are 30° and 120°

Solution. It follows from the assumptions that the center line of the AB segment with the vertices C and D divides the ABCD trapezoid into three identical equilateral triangles. Therefore, the magnitude of internal angles in the trapezoid at A and B vertices is equal to 60 °

And at the C and D vertices 120 °. It follows from the specification that the triangles LCK and MDL are the same (according to the sentence above). Therefore, both the KL and LM lines and the designated pairs of angles are the same; The magnitudes of these angles are denoted α and β. The triangle KLM is isosceles and the angles at the base are the same; Their size is denoted by δ and the size of the angle KLM is denoted by γ.

From the sum of the inner angles in the KCL triangle we derive

α + β = 180° − 120° = 60°

The sum of the three marked angles with the vertex L is a straight angle, therefore

γ = 180° − (α + β) = 120°

Finally, we deduce the sum of inner angles in the triangle KLM

δ = (180° − 120°)/2 = 30°

The internal angles of the triangle KLM are 30° and 120°

#### To solve this example are needed these knowledge from mathematics:

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