This page shows how to construct a right triangle that has the hypotenuse (H) and one leg (L) given. It is almost the same construction as Perpendicular at a point on a line, except the compass widths used are H and L instead of arbitrary widths.
The above animation is available as a printable step-by-step instruction sheet, which can be used for making handouts or when a computer is not available.
It is possible to draw more than one triangle has the side lengths as given. Youcan use the triangle to the left or right of the initial perpendicular, and also draw them below the initial line. All four are correct in that they satisfy the requirements, and are congruent to each other.
This construction works by effectively building two congruent triangles. The image below is the final drawing above with the blue line BP added
|We first prove that ∆BCA is a right triangle|
|1||CP is congruent to CA||They were both drawn with the same compass width|
|2||BP is congruent to BA||They were both drawn with the same compass width|
|3||CB is common to both triangles BCP and BCA||Common side|
|4||Triangles ∆BCP and ∆BCA are congruent||Three sides congruent (SSS).|
|5||∠BCP, ∠BCA are congruent||CPCTC. Corresponding parts of congruent triangles are congruent|
|6||m∠BCA = 90°||∠BCA and ∠BCP are a linear pair and (so add to 180°) and congruent so each must be 90°|
|We now prove the triangle is the right size|
|7||CA is congruent to the given leg L||CA copied from L. See Copying a segment.|
|8||AB is congruent to the given hypotenuse H||Drawn with same compass width|
|9||∆BCA is a right triangle with the desired side lengths||From (6), (7), (8)|