Method 1.
Look at the picture 1.
We have:
[tex]2\alpha=180^o\ \ \ \ |:2\\\\\alpha=90^o[/tex]
Picture 2.
[tex]m\angle LMN=90^o[/tex]
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Method 2. (Picture 3)
ΔLOM and ΔMON are isosceles triangles. Therefore the two angles opposite the legs are equal.
The sum of the internal angles in each triangle is 180°. Thereofre in ΔLMN we have:
[tex]\alpha+\beta+\beta+\alpha=180^o\\\\2\alpha+2\beta=180^o\ \ \ |:2\\\\\alpha+\beta=90^o[/tex]
[tex]m\angle LMN=\alpha+\beta\to m\angle LMN=90^o[/tex]
