007. Types of the Triangle

Let MM be the midpoint of BC\overline{BC} of a triangle ABCABC. Determine the types of this triangle if BAM+ACB=90°\angle BAM + \angle ACB = 90\degree.

First, note that C\angle C cannot be an obtuse angle or right angle to satisfy the condition. Besides, B\angle B cannot be an obtuse angle or right angle as well because BAMCAH\angle BAM \ne \angle CAH where HH is the perpendicular from AA to the line BCBC.


Therefore, HH must be on BC\overline{BC} except for BB and CC. The following two triangles satisfy BAM=CAH\angle BAM = \angle CAH and can become candidates.


Now, let DD be the intersection between the circumcircle of ABCABC and the line AMAM, which is not AA. Then, by inscribed angle theorem, ACB=ADB\angle ACB = \angle ADB and ABD=90°\angle ABD = 90 \degree. That is, AD\overline{AD} is a diameter of this circumcircle.


Accordingly, since AD\overline{AD} passes through the midpoint MM, BC\overline{BC} is a diameter or BCAD\overline{BC} \perp \overline{AD}.


Therefore, the triangle ABCABC are a right triangle with its right angle at AA or an isosceles triangle such that AB=AC\overline{AB} = \overline{AC}.

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