Question

Given: Circle O with diameter LN and inscribed angle LMN Prove: is a right angle.

Answer 1

Method 1.

Look at the picture 1.

We have:

$2\alpha=180^o\ \ \ \ |:2\\\\\alpha=90^o$

Picture 2.

$m\angle LMN=90^o$

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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:

$\alpha+\beta+\beta+\alpha=180^o\\\\2\alpha+2\beta=180^o\ \ \ |:2\\\\\alpha+\beta=90^o$

$m\angle LMN=\alpha+\beta\to m\angle LMN=90^o$