Question

right triangle abc is isosceles and point m is the midpoint of the hypotenuse.
what is true about triangle amb?
it is congruent to triangle abc.
it is an isosceles right triangle.
it is an obtuse triangle.
it is a scalene triangle.

Explanation:

Step1: Analyze triangle ABC

Triangle \( ABC \) is a right isosceles triangle, so \( AB = BC \) and \( \angle B = 90^\circ \). \( M \) is the midpoint of hypotenuse \( AC \), so \( AM = MC \). Also, in a right - isosceles triangle, the median to the hypotenuse is half the hypotenuse, so \( BM = AM = MC \).

Step2: Analyze triangle AMB

- For the first option: Triangle \( AMB \) and \( ABC \): \( ABC \) has sides \( AB, BC, AC \) with \( AB = BC \) and \( AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{2}AB \). \( AM=\frac{AC}{2}=\frac{\sqrt{2}AB}{2}\), \( AB = AB \), \( BM = AM=\frac{\sqrt{2}AB}{2}\). The sides of \( AMB \) are \( AB, AM, BM \) with \( AM = BM \), and \( \angle ABM \): Since \( ABC \) is right - isosceles and \( M \) is the midpoint, \( \angle ABM = 45^\circ\), \( \angle BAM=45^\circ \), so \( \angle AMB = 90^\circ \)? Wait, no. Wait, \( ABC \) is right - isosceles with \( \angle B = 90^\circ \), \( AB = BC \). \( M \) is the midpoint of \( AC \), so \( BM\) is the median to the hypotenuse. In a right triangle, the median to the hypotenuse is equal to half the hypotenuse, so \( AM = BM = MC \). Also, since \( AB = BC \) and \( M \) is the midpoint, triangle \( AMB \): \( AB \) is a leg of the original right triangle, \( AM = BM \) (because \( AM=\frac{AC}{2}\) and \( BM=\frac{AC}{2}\)), and \( \angle AMB \): Let's check the angles. \( \angle BAC = 45^\circ \) (since \( ABC \) is isosceles right - triangle), \( AM = BM \), so triangle \( AMB \) has \( AM = BM \) and \( \angle MAB=45^\circ \), so \( \angle MBA = 45^\circ \), then \( \angle AMB=180 - 45 - 45=90^\circ \)? Wait, no, maybe I made a mistake. Wait, the original triangle \( ABC \) is right - isosceles with right angle at \( B \). \( M \) is the midpoint of \( AC \). So \( AB = BC \), \( AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{2}AB \), \( AM = MC=\frac{\sqrt{2}AB}{2}\), \( BM=\frac{AC}{2}=\frac{\sqrt{2}AB}{2}\) (property of right - triangle: median to hypotenuse is half the hypotenuse). So in triangle \( AMB \), \( AM = BM=\frac{\sqrt{2}AB}{2}\) and \( AB \) is a side. So two sides \( AM = BM \), so it is isosceles. Also, let's check the angle at \( B \) in triangle \( AMB \). The angle \( \angle ABM \): In triangle \( ABC \), \( \angle ABC = 90^\circ \), and \( BM \) is the median. In an isosceles right - triangle, the median to the hypotenuse also bisects the right angle? Wait, no, in an isosceles right - triangle with right angle at \( B \), the median from \( B \) to \( AC \) would bisect \( \angle B \), but here \( M \) is the midpoint of \( AC \), so \( BM \) is the median from \( B \) to \( AC \). Wait, maybe I confused the median. Wait, the triangle \( AMB \): \( AB \) is equal to \( BC \), \( AM = BM \), and \( \angle AMB \): Let's use the side - angle - side. \( AB = BC \), \( AM = BM \), \( \angle BAM=\angle BCM = 45^\circ \)? No, \( \angle BAC = 45^\circ \), since \( ABC \) is isosceles right - triangle. So in triangle \( AMB \), \( AM = BM \), so it is isosceles. Also, is it a right triangle? Let's check the angles. If \( AM = BM \) and \( \angle MAB = 45^\circ \), then \( \angle MBA=45^\circ \), so \( \angle AMB = 90^\circ \). Wait, but the option says "It is an isosceles right triangle". Let's check other options:
- "It is congruent to triangle \( ABC \)": The sides of \( ABC \) are \( AB, BC, AC \) with \( AB = BC \), \( AC=\sqrt{2}AB \). The sides of \( AMB \) are \( AB, AM, BM \) with \( AM = BM=\frac{\sqrt{2}AB}{2}\). So the sides are not equal, so not congruent.
- "It is an obtuse triangle": Since we found that \( \angle AMB = 90^\circ \) (if our previous calculation is correct) or at least the angles are \( 45^\circ,45^\circ,90^\circ \), so it's not obtuse.
- "It is a scalene triangle": Since \( AM = BM \), it has two equal sides, so not scalene.
- "It is an isosceles right triangle": Since \( AM = BM \) (two equal sides, so isosceles) and \( \angle AMB = 90^\circ \) (right angle), this option is correct.

Answer:

It is an isosceles right triangle.