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Home » Triangles » Congruent Triangles » HL Theorem (Hypotenuse-Leg Postulate)

HL Theorem (Hypotenuse-Leg Postulate)

Last updated: Mar 27, 2021 by Ido Sarig · This website generates income via ads and uses cookies · Terms of use · Privacy policy

The HL theorem (with HL standing for Hypotenuse-Leg) is sometimes used to describe a fifth postulate for proving that triangles are congruent, in addition to the 4 basic ones (Side-Side-Side, Side-Angle-Side, Angle-Side-Angle, and Angle-Angle-Side).

The HL theorem only applies to right triangles, as the 'hypotenuse' specification implies. So to use it we need (1) a right angle, (2) a congruent side which is opposite that right angle - the hypotenuse, and (3) one of the other two sides.

We have already proven this theorem, using Pythagorean theorem. But here, we'll prove the HL theorem a different way, which does not rely on the advanced concept of Pythagorean theorem.

Problem: Prove the HL Theorem

Two right triangles, ΔABC and ΔDEF have an equal hypotenuse (|AC|=|DF|) and an equal leg (|BC|=|EF|). Prove that the triangles are congruent.

Congruent right triangles - HL Theorem

Strategy

We already have a couple of congruent sides, and a congruent angle, so we'll try to see if we can use one of the other 3 congruent postulates, by constructing another congruent side or angle.

Let's take triangle ΔABC, extend edge AB with a straight edge, and using a compass, construct a point G so that |BG|=|DE|.

To make it easier to see what we are doing, let's color-code the congruent sides:

Color coded triangles

Now, consider triangles ΔDEF and ΔGBC: |BG|=|DE| by construction (the green leg), and |BC|=|EF|, as given in the problem formulation (the red leg).

∠CBG is a right angle, since it forms a linear pair with ∠CBA, which is a right angle as given in the problem. So m∠CBG = m∠DEF = 90° and ΔDEF ≅ ΔGBC by the Side-Angle-Side postulate.

If ΔDEF ≅ ΔGBC, then |CG|=|DF|, as corresponding sides of congruent triangles, and since |AC|=|DF| (given), |CG|=|AC| (transitive property of equality), making ΔACG an isosceles triangle. We have shown that in an isosceles triangle, the height to the base is also the median, so |AB|=|BG|. But |BG|=|DE| as corresponding sides in congruent triangles (ΔDEF ≅ ΔGBC), so |AB|=|DE| and we now have our third congruent side , and (ΔDEF ≅ ΔABC) by Side-Side-Side.

Proof

(1) |AC|=|DF| //Given
(2) |BC|=|EF| //Given
(3) m∠CBA =90° //Given
(4) m∠DEF =90° //Given
(5) |BG|=|DE| //Construction
(6) m∠CBG = m∠DEF = 90° //(3), ∠CBG and ∠CBA are a linear pair
(7) ΔDEF ≅ ΔGBC //(2), (5), (6), Side-Angle-Side
(8) |CG|=|DF| // corresponding sides of congruent triangles (CPCTC)
(9) |AC|=|DF| =|CG| //(1) , (8)
(10) |AC|=|CG| //(9) , transitive property of equality
(11) |AB|=|BG| //(10), ΔACG an isosceles triangle, (6) height to base of isosceles triangle is the median
(12) |AB|=|DE| //(11), (5) transitive property of equality
(13) ΔDEF ≅ ΔABC //(2), (1), (12), Side-Side-Side

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About the Author

Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. His goal is to help you develop a better way to approach and solve geometry problems. You can contact him at [email protected]

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Welcome to Geometry Help! I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree in Management of Technology. I'm here to tell you that geometry doesn't have to be so hard! My goal with this website is to help you develop a better way to approach and solve geometry problems, even if spatial awareness is not your strongest quality. Read More…

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