The cases where AB=AC (blue), AB=BC (red), and AC=BC (green) (lighter versions on the left side of the y-axis) are shown below for measures of angle C between 0 and 180°. The altitude drawn from the vertex angle to the base divides an isosceles triangle into two congruent right triangles. Mangaldan's comment), and by symmetry this perpendicular bisector of AB also bisects $\angle ACB$ from there, you can use right triangle trigonometry to determine the coordinates of C (left for you to solve). Side opposite the vertex angle is the base. The height divides the base into two equal parts, creating two right triangles. The point M lies on C H because the triangle is isosceles, so in A B M, M H is a median and an altitude so its an isosceles triangle. To find the height of an isosceles triangle, you can use the Pythagorean theorem. Let C H be the third median of the triangle. In the third case, C is equidistant from A and B, so C must lie on the perpendicular bisector of AB (as in J. The area of an isosceles triangle is S and the angle between the medians to the legs, facing the base, is. The second case is similar to the first (so it's left for you to solve). By using the previous formula, these values for the area and the base can be plugged in so the height can be determined. Edit (to match revised question): Given your revised question, there is still the issue of C being on either side of the y-axis, but you have specified that AB=AC and that you are given $\mathrm\angle C))$. The formula h ( a2b2/4) is used as a calculation tool to determine the altitude of an isosceles triangle.
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