Theory of right triangles

T34 If the hypotenuse and a leg of one right triangle are congruent to the corresponding parts of another right triangle, then the triangles are congruent.

[Maple OLE 2.0 Object]

Given: right triangles ABC and DEF with hypotenuses AB = DE and legs AC = DF

To show : angle(A) is congruent with angle(D) and hence by SAS triangle(ABC) is congruent with triangle(DEF).

Proof outline . Suppose not. We can assume that m(angle(A)) < m(angle(D)) (why?) Then construct B' on FE

so that m(angle(FDB')) = m(angle(A)). So triangle CAB is congruent with triangle FDB' (why?). So AB = DB' (why?). So m(angle(EB'D))= m(angle(B'ED)) (why?) So m(angle(EB'D)) < 90 (why?). But m(angle(EB'D)) > 90 by the exterior angle theorem. qed

Note: There is a much shorter proof using pythagoras' theorem.

T35 The median to the hypotenuse of any right triangle is half as long as the hypotenuse.

Idea of proof . Drop a perpendicular from the midpoint of the hypotenuse to one of the legs. Establish that the right triangle cut off is similar to the original by a factor of 1/2. Then show that the right triangle cut off is congruent to the right triangle whose hyponuse is the median to the hypotenuse of the original triangle.

T36 In any right triangle, the altitude to the hypotenuse forms two right triangles that are similar to each other and to the original triangle.

[Maple OLE 2.0 Object]

Hint on Proof: Use the labelled diagram (which theorems do you need to justify the two angles marked with the same number are congruent?)

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T37 Given a right triangle, the altitude to the hypotenuse divides the hypotenuse into two segments such that the altitude is their geometric mean.

Hint : Call the lengths of the two segments c1 and c2, and call the lenght of the altitude h. Use theorem 36 to set up a proportion to derive the result h = sqrt(c1*c2) .

T38 The square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of its legs.

Hint : Make a labelled diagram. Use T 36 to write down several proportions. From some of these derive the pythagorean equation.

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Here is a graph of the theorems up to T38. Quite a cobweb when they are all put together, isn't it?

T39 In a 30 60 degree right triangle, the short leg is half the hypotenuse.

Hint: Reflect the triangle about its long leg (the one opposite the 60 degree angle). See an equilangular triangle anywhere?

T40 For any angle, the sum of the squares of its sine and cosine is 1.

Hint: Use the definition of sine and cosine of an angle and pythagoras' theorem.

T41 In any triangle, the ratio of the length of a side to the sine of the angle opposite it is the same for all three sides.

[Maple OLE 2.0 Object]

Hint on proof. Use this labelled diagram. Write b sin(A) = h = a sin(B) and try a little algebra.

T42 . In any triangle, the square on the length of a side is the sum of the squares on the lengths of the other two side reduced by twice the product of their lengths and the cosine of the angle between them.

Hint: Make a labelled diagram.

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