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Hi ! I have some problem with this below !
Subjected to a square which has two isosceles triangles.
One of them is: a triangle whose base is one of the sides of the square with base angles of 15.
The other is a triangle whose base is the side opposite the base of the triangle it first touches the apex of the triangle at the end of the first node.
Proved that the second triangle equilateral!
The prove must be geometric way , no trigonometric ! Please help me ! thanks all !
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Draw square BCED and circle, centre B, radius BD (=BC).
Construct line DA so that angle EDA = 15, and A lies on the circle.
ADB = 90 15 = 75
Triangle ADB is isosceles as AB = BD = radius.
So, BDA = BAD = 75, and therefore ABD = 30.
Therefore, ABC = 60.
But BA = BC (= radius) so triangle BAC is isosceles,
So BAC = BCA = (180-60)/2 = 60
So triangle BAC is equilateral.
Draw a line of symmetry through A, parallel to DB,
CAE is another isosceles triangle (75,75,30)
So ADE is the 15, 15 150 triangle.
Last edited by Bob (2010-08-09 06:00:31)
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