Question 1175969
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Angle B is a 60-degree angle by virtue of the fact that it is a vertex of an equilateral triangle and therefore is the vertex of an equiangular triangle whose measure must be one-third of 180 degrees.


Since angle BYZ is right, angle BZY must be 30 degrees (180 - 90 - 60 = 30)


Therefore triangle BYZ is a 30-60-90 right triangle, and by similar logic and the fact that triangle XYZ is equilateral as a given, triangles BYZ, AXY, and CZX are all congruent.  Congruency demands that the three triangles mentioned must be equal in area.


Let the measure of segment BZ be 1 unit.  Then by the properties of a 30-60-90 right triangle, segments BY, CZ, and AX must be *[tex \Large \frac{1}{2}] unit, and YZ, ZX, and XY must be *[tex \Large \frac{\sqrt{3}}{2}] units.


Since segment BC is the sum of segments BZ and ZC, the measure of BC must be *[tex \Large \frac{3}{2}] units, so if you drop a perpendicular to AC from vertex B at M, triangle BCM is both a 30-60-90 right triangle and is one-half of the area of triangle ABC.  By the properties of a 30-60-90 right triangle, *[tex \Large MC\ =\ \frac{3}{4}] units and *[tex \Large BM\ =\ \frac{3\sqrt{3}}{4}]


Area of triangle ABC: *[tex \Large 2\(\frac{BM * MC}{2}\)\ =\ 2\(\frac{\(\frac{3 \sqrt{3}}{4}\)\(\frac{3}{4}\)}{2}\)\ =\ \frac{9 \sqrt{3}}{16}  ] units².


Area of triangle BYZ: *[tex \Large \frac{\(\frac{1}{2}\)\(\frac{\sqrt{3}}{2}\)}{2}\ =\ \frac{\sqrt{3}}{8}].  And the sum of the areas of triangles BYZ, AXY, and CZX is 3 times the area of BYZ, to wit: *[tex \Large \frac{3 \sqrt{3}}{8}]


The area of triangle XYZ is the difference between the area of triangle ABC and the sum of the areas of the three triangles just calculated, namely *[tex \Large \frac{9 \sqrt{3}}{16}\ -\ \frac{3 \sqrt{3}}{8}\ =\ \frac{3 \sqrt{3}}{16}]


Then the ratio of the area of XYZ to ABC is *[tex \Large \frac{\frac{3 \sqrt{3}}{16}}{\frac{9 \sqrt{3}}{16}}\ =\ \frac{1}{3}]

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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