Question

geometry with data analysis ic qtr b young continuous learning center - credit bearing (tutor)
isosceles triangles
right triangle abc is isosceles and point m is the midpoint of the
what is true about triangle amb?
hypotenuse.
it is congruent to triangle abc.
it is an isosceles right triangle.
it is a scalene triangle.
it is an obtuse triangle.
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Explanation:

Step1: Recall properties of isosceles right - triangle and mid - point of hypotenuse

In an isosceles right - triangle \(ABC\) with \(\angle B = 90^{\circ}\) and \(AB=BC\), and \(M\) is the mid - point of the hypotenuse \(AC\). By the property of a right - triangle, \(AM = BM=CM=\frac{1}{2}AC\).

Step2: Check the angles and side lengths of \(\triangle AMB\)

Since \(AB = BC\) and \(AM = BM\) (from the property of the mid - point of the hypotenuse of a right - triangle \(AM = BM=\frac{1}{2}AC\)), and \(\angle ABM = 45^{\circ}\), \(\angle BAM=45^{\circ}\), \(\angle AMB = 90^{\circ}\) (because the median to the hypotenuse of a right - triangle has some angle relationships. Also, using coordinate geometry or vector methods (if we assume \(B=(0,0)\), \(A=(0,a)\), \(C=(a,0)\), then \(M = (\frac{a}{2},\frac{a}{2})\), \(AM=\sqrt{(\frac{a}{2}-0)^{2}+(\frac{a}{2}-a)^{2}}=\frac{\sqrt{2}a}{2}\), \(BM=\sqrt{(\frac{a}{2}-0)^{2}+(\frac{a}{2}-0)^{2}}=\frac{\sqrt{2}a}{2}\), \(AB = a\), \(\angle AMB = 90^{\circ}\))

A congruent triangle would have all sides equal. Here \(AM=\frac{1}{2}AC\), \(AB
eq AM\) (since in right - isosceles triangle \(AC=\sqrt{2}AB\)), so \(\triangle AMB\) is not congruent to \(\triangle ABC\). A scalene triangle has all sides of different lengths, but \(AM = BM\). An obtuse triangle has one angle greater than \(90^{\circ}\), but \(\angle AMB=90^{\circ}\)

Answer:

It is an isosceles right triangle.