Ninth Grade Mathematics: RHS (Hypotenuse-Leg) Criterion

Right-angled triangles possess unique geometric properties governed by the Pythagorean theorem. While general Side-Side-Angle (SSA) is invalid because of the ambiguous case, fixing the angle at exactly 90 degrees eliminates all ambiguity. The Right-angle-Hypotenuse-Side (RHS) criterion, also known as the Hypotenuse-Leg (HL) theorem, provides the definitive congruence test for right triangles.

Statement and Meaning of the RHS Criterion

The RHS criterion states:

The RHS (Right-angle-Hypotenuse-Side) Congruence Criterion:
Two right-angled triangles are congruent if the hypotenuse and one leg
of one triangle are equal to the hypotenuse and one leg of the other triangle.

               TRIANGLE ABC                       TRIANGLE DEF
                    A                                  D
                    *                                  *
                    | \                                | \
                    |  \                              |  \
                  b |   \ c                         b |   \ c
                    |    \                            |    \
                    *-----+                            *-----+
                    B  a  C                            E  a  F
                    Angle B = Angle E = 90 degrees (Right Angle)
                    Hypotenuse AC = Hypotenuse DF (Hypotenuse)
                    Leg AB = Leg DE (Side / Leg)

Notice that although this appears to be a Side-Side-Angle configuration (with sides c and b and non-included angle B = 90 degrees), the presence of the right angle guarantees that only one single right triangle can ever be constructed.

Proof of the RHS Criterion via the Pythagorean Theorem

We can prove the RHS criterion algebraically using the Pythagorean theorem and the SSS criterion:

Given: Triangle ABC and Triangle DEF such that:
       Angle B = Angle E = 90 degrees,
       Hypotenuse AC = Hypotenuse DF = h,
       Leg AB = Leg DE = l.
To Prove: Triangle ABC =~ Triangle DEF.

Proof:
Step 1: Apply the Pythagorean theorem to right-angled Triangle ABC:
        AB^2 + BC^2 = AC^2
        l^2 + BC^2 = h^2
        BC^2 = h^2 - l^2
        BC = sqrt( h^2 - l^2 )

Step 2: Apply the Pythagorean theorem to right-angled Triangle DEF:
        DE^2 + EF^2 = DF^2
        l^2 + EF^2 = h^2
        EF^2 = h^2 - l^2
        EF = sqrt( h^2 - l^2 )

Step 3: Comparing the two results:
        BC = EF.

Step 4: We now have:
        AB = DE (Given)
        AC = DF (Given)
        BC = EF (Proven from Pythagorean theorem)

Step 5: By the SSS Congruence Criterion:
        Triangle ABC =~ Triangle DEF.

Because the third side of any right triangle is uniquely determined by its hypotenuse and one leg, the RHS criterion is mathematically equivalent to the SSS criterion.

Why the Ambiguous Case Disappears at 90 Degrees

Recall that the ambiguous case of SSA occurs when an arc swings from the top vertex and cuts the baseline in two locations. However, the shortest distance from any point to a line is the perpendicular segment:

              A
              *
              | \
              |  \
            h |   \ Hypotenuse c
              |    \
              +-----*-------------------
              B     C
            90 deg

At 90 degrees, leg AB is the perpendicular line. Any hypotenuse segment c longer than AB can only cut the horizontal baseline at a single point on either side of B. Thus, there is exactly one possible right triangle, completely eliminating the ambiguity.

Chapter Exercises

Exercise 1: In triangle ABC, altitudes AD and BE are drawn to sides BC and AC respectively such that AD = BE. Prove that triangle ABC is an isosceles triangle using the RHS congruence criterion.

Exercise 2: Point P is equidistant from two intersecting lines l and m. Prove that ray OP bisects the angle formed between lines l and m using RHS congruence.

Exercise 3: Two right triangles ABC and DBC share the common hypotenuse BC and lie on opposite sides of BC. If AB = DC, prove that AC = DB and Angle BAC = Angle BDC.

Solutions and Explanations

Solution 1: Consider the right-angled triangles BEC and ADC. Because AD and BE are altitudes, Angle BEC = Angle ADC = 90 degrees (Right angle). Both triangles share the common hypotenuse BC (Hypotenuse BC = BC). We are given that the altitudes are equal: BE = AD (Leg). By the RHS Congruence Criterion, Triangle BEC =~ Triangle ADC. By CPCTC, their corresponding angles are equal: Angle BCE = Angle ACD, which means Angle C = Angle C, but more importantly, Angle B = Angle C. In triangle ABC, since two angles (Angle B and Angle C) are equal, the sides opposite them are equal: AB = AC. Therefore, triangle ABC is an isosceles triangle.

Solution 2: Let lines l and m intersect at vertex O. From point P, draw perpendiculars PA to line l and PB to line m. Since P is equidistant from the lines, PA = PB. Consider right triangles OAP and OBP. Angle OAP = Angle OBP = 90 degrees (Right angle). The line segment OP is the hypotenuse common to both triangles (Hypotenuse OP = OP). We are given PA = PB (Leg). By the RHS Congruence Criterion, Triangle OAP =~ Triangle OBP. By CPCTC, corresponding angles at the vertex O are equal: Angle AOP = Angle BOP. This proves that ray OP bisects the angle between lines l and m.

Solution 3: Consider right triangles ABC and DCB. Both triangles have a right angle, with common hypotenuse BC (Hypotenuse BC = BC). We are given that side AB = side DC (Leg). By the RHS Congruence Criterion, Triangle ABC =~ Triangle DCB. By CPCTC, the corresponding remaining sides are equal: AC = DB. Furthermore, by CPCTC, their corresponding angles are equal: Angle BAC = Angle BDC (both are the right angles, or corresponding angles). Thus, both statements are proven.