Problem
GEO-B1-M07-P023 A Median as an Altitude
#23
★★★★☆ Level 4 of 5
In triangle \(ABC\), point \(M\) is the midpoint of \(BC\). It is known that \(AM\perp BC\). Prove that \(AB=AC\). What construction or comparison is natural here?
Compare triangles \(ABM\) and \(ACM\).
Triangles \(ABM\) and \(ACM\) are right triangles. They have \(BM=CM\), since \(M\) is the midpoint, and \(AM\) is common. Thus the triangles are congruent by two legs, so \(AB=AC\). The natural action here is to look for congruent triangles around the median.
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