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|dw:1338413561449:dw| construct equilateral triangle external to the square. Sides AE= AD (sides of square)
|dw:1338413811564:dw| triangle AEP congruent to triangle ADP by side AP, angle 75, side AD = side AE triangle AEP is isosceles (75 - 30 - 75) so triangle ABD is also, and AD= DP by similar argument using triangles BEP and BCP, side CP = side BC. As CD=AD=BC (sides of a square), all three sides PD,CD and CP are equal and triangle CDP is equilateral
suppose that the figure in bottom occurs?|dw:1338509358894:dw|
Triangle ABE is equilateral by construction. You can always construct it. triangle APB is isosceles, so points E and P will lie on the perpendicular bisector.
Ok but you suppose that AD=AE so you can not assume these assumptions together ?
You construct equilateral triangle AEB. One side of it is AB. AB = AD (sides of a square). As all 3 sides are equal we know AE= AB, and therefore AE= AD. triangle APB is isosceles (given) triangle AEB is equilateral (by construction) therefore, we can show the perpendicular dropped from E to AB will bisect AB. Similarly, the perpendicular dropped from P to AB will bisect AB. therefore PE is the perpendicular bisector of AB
we can show the perpendicular dropped from E to AB will bisect AB. Given isosceles triangle ABE with angle A= angle B |dw:1338512498308:dw| because 2 of the three angles are equal, the 3rd angle must be equal so by angle, side (the perpendicular), angle (the right angle), the 2 triangles are congruent, and so the perpendicular is also the bisector.
I think you did not get my question. Please look at the picture and explain how you suppose ED is one straight line? If you assume two sides has the same length you can not suppose ED is a line which Perpendicular in AB.
I got your question. there is no gap.
also like this we do not know!|dw:1338515217497:dw|