Question 698039
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Two ways to go about it.


1.  Construct the second triangle adjacent to the first so that the two legs are coincident.  This produces an isosceles triangle where the hypotenuse of the first triangle is one of the congruent sides of the isosceles triangle and the hypotenuse of the other is the other congruent side.  The coincident legs are the altitude.  There is a theorem (look it up) that says the altitude to the vertex of an isosceles triangle divides the triangle into two congruent triangles.


2.  Use the Pythagorean theorem.  If you have two right triangles with sides a, b, c (c the hypotenuse) and d, e, f (f the hypotenuse) and you know that a = d and c = f, then with Pythagoras you can show that *[tex \LARGE b^2\ =\ c^2\ -\ a^2] and *[tex \LARGE e^2\ = f^2\ -\ d^2].  But by substitution, *[tex \LARGE e^2\ =\ c^2\ -\ a^2].  So *[tex \LARGE e^2\ =\ b^2] which implies that *[tex \LARGE e\ =\ b].  Therefore the triangles are congruent by SSS.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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