Theorem: In a right triangle, the hypotenuse squared is equal to the sum of the other sides squared.

Proof:

(1) Since BAC is a right angle, we can extend AC to make a square GFBA and we can extend AB to make a square ACKH. (See here for details on the construction)

(2) We can also build a square based on BC. (See here for details on the construction)

(3) Let AL be a line that is parallel to BD and CE. (See here for details on the construction)

(4) ∠ DBA ≅ ∠ FBC since ∠ FBA and ∠ DBC are right angles and we get this result if we add ∠ ABC to both.

(5) triangle DBA ≅ triangle FBC by Side-Angle-Side (see here) since:

(a) ∠ DBA ≅ ∠ FBC (#4)

(b) FB ≅ AB since they are sides of the same square.

(c) BC ≅ BD since they are sides of the same square.

(6) The parallelogram BL is double the triangle ABD since they have the same base (BD) and since they are in the same parallel. (see Corollary 3.1 here for details)

(7) Likewise, the parallelogram GFBA is double the triangle FBC for the same reason as step #6.

(8) From step #5, we can conclude that the parallelogram BL is congruent to the parallelogram GFBA.

(9) We can follow the same steps to show that parallelogram CL is congruent to parallelogram ACKH since:

(a) ∠ KCB ≅ ∠ ACE since ∠ BCE, ∠ ACK are right angles and we can add ∠ ACB to each.

(b) AC ≅ CK and BC ≅ CE since the sides of the same square are congruent.

(c) So that triangle KCB ≅ triangle ACE by Side-Angle-Side.

(d) parallelogram CL is double the area of triangle ACE

(e) parallelogram CA is double the aera of triangle KCB

(f) So therefore parallelogram CL ≅ parallelogram CA

(10) So that we have the square of BC is congruent to the square of AB added to the square of AC since:

(a) the square BDEC = parallelogram BL + parallelogram CL

(b) parallelogram BL ≅ square of AB

(c) parallelogram CL ≅ square of AC

QED

Corollary 1: In a right triangle, the hypotenuse is longest of the three sides of a triangle.

Proof:

(1) Assume that the hypotenuse h is equal or less than one side s

_{1}

(2) Then, h

^{2}is less than s

_{1}

^{2}+ s

_{2}

^{2}

(3) But this contradicts the Pythagorean Theorem so we can reject our assumption.

QED

Corollary 2: sin

^{2}(θ) + cos

^{2}(θ) = 1.

Proof:

(1) From the main theorem, for any right triangle with base sides x and y and with hypotenuse z, we have:

x

^{2}+ y

^{2}= z

^{2}

(2) If θ is the angle between z and x, then we have (see here):

sin θ = y/z

cos θ = x/z

So that:

sin

^{2}(θ) = y

^{2}/z

^{2}

cos

^{2}(θ) = x

^{2}/z

^{2}

(3) Now if we divide the equation in step #1, by z

^{2}on both sides we get:

x

^{2}/z

^{2}+ y

^{2}/z

^{2}= 1.

(4) Now inserting the values in step #2 gives us:

sin

^{2}(θ) + cos

^{2}(θ) = 1.

QED