PROPOSITION: The hypotenuses of two dissimilar right triangles “A” and “B”, are twice the legs of a known right triangle “C”, and the altitude to hypotenuses in each of A and B triangles are identical to that in C.

HYPOTHESIS: The sum of the greater segments on each of the hypotenuses of A and B, caused by the altitude to hypotenuse in C, will equal the Perimeter of C.

For instance, triangle C can be any right triangle whose 3 sides are known.

Let “m” and “n” be the long and short legs, respectively, of C.

Let "d" be the altitude to hypotenuse of C.

Let "h" be the hypotenuse of C.

Then, the hypotenuses of A and B will equal 2(m) and 2(n), respectively, and the altitude to their hypotenuses will be "d", same as that in C.

**Correlation**: Two right triangles A and B, with hypotenuses equal to twice the legs of a known right triangle C, and the same altitude to the hypotenuse as in C, will contain acute ∠’s equal to one half the acute ∠’s in C.

The lesser acute ∠ in the triangle with the greater hypotenuse (2m), for triangle **A**, will be 1/2 the smaller acute ∠ in C.

The lesser acute ∠ in the triangle with the smaller hypotenuse (2n), for triangle **B**, will be 1/2 the greater acute ∠ in C.

Pythagorean Triangles are suggested for this problem to facilitate ease of calculations.

By the Proposition in Post 1, right triangle C in this example is the 3 4 5 Pythagorean triangle, whose altitude to hypotenuse "d", is 2.40, and Perimeter is 12.00.

☆The greater segments on the hypotenuses of right triangles A and B are: 7.20 for A, and 4.80 for B. See calculations below:

Hyp of A = 2m = 2(4) = 8.00.

Hyp of B = 2n = 2(3) = 6.00.

The lesser∠ in A =18.43494882° = 1/2 (36.86989765°) <∠ in C.

2.40 /Tan 18.43494882° = 7.20, >Seg on Hyp of

A.The lesser∠ in B = 26.56505118° = 1/2 (53.13010235°) >∠ in C.

2.40 /Tan 26.56505118° = 4.80, >Seg on Hyp of

B.Perimeter of A = 7.20 + 4.80 = 12.00.

For any right triangle C, the known triangle, the product of the hypotenuses of A and B divided by the Area of C will always be 8.00.

In this Example C is the 5 12 13 triangle. Per the Proposition, hypotenuse of A = 24.00, or 2 (12), and hypotenuse of B = 10.00, or 2(5). So, Area of C= 30.00, and 24.00 x 10.00 / 30.00 = 8.00.

By the Proposition in Post 1, we have the following Conjecture: The long leg of B equals the short leg of B multiplied by the long leg of C, (divided by the hypotenuse of C minus the short leg of C).

EXAMPLE:

Using the 5 12 13 triangle for C, we solve for the product of 2n (n = short leg of C), and 5.00, for the hypotenuse of B, to be10.00.

NOTE:.The altitude to hypotenuse in B is same as that in C, but its value is not required for this conjecture, however, it makes the conjecture possible.

> ∠ in C = 67.38013505°.

< ∠ in B = 1/2 > ∠ in C = 1/2 (67.38013505°) = 33.69006753°.

Short leg of B = 10 x Sin 33.69006753° = 5.547001962

12 x 5.547001962 = 66.56402354.

Long leg of B = 66.56402354/(13-5)= 8.320502943.

Long leg of B = 10 x Cos 33.69006753° = 8.320502943

Problem: The sum of the areas of right triangles A and B, divided by the altitude to the hypotenuse in right triangle C is equal to the sum of the legs of C.Right triangle C in Post 3 was the 3 4 5 triangle. Altitude to hypotenuse in C is 2.40.

Per Post 1, Altitude to hypotenuse in A and B is also 2.40.

Let X = 2.40

Hyp of A = 2(4) = 8.00.

Hyp of B = 2(3) = 6.00.

Area of A = 8X / 2 = 4X = 9.60.

Area of B = 6X / 2 = 3X = 7.20.

9.60 + 7.20 / 2.40 = 16.80 / 2.40 = 7.00.

Areas of A plus B, divided by the altitude to hypotenuse.

4X + 3X / X = 7X / X = 7 = 4 + 3, sum of legs of triangle C.