Additional congruence criteria (RHS)

RHS congruence — stated without proof. If, in two right triangles, the hypotenuse and one side of one triangle are equal, respectively, to the hypotenuse and the corresponding side of the other triangle, then the two triangles are congruent. Write it for $\triangle ABC$ and $\triangle DEF$, both right-angled at $B$ and $E$, with hypotenuses $AC$ and $DF$:

where: $AC$ and $DF$ are the hypotenuses (each opposite its triangle's right angle), $AB$ and $DE$ are one corresponding pair of legs, and $\angle B = \angle E = 90°$ is the right angle common to both triangles, already known to be equal before RHS is even invoked.

Why the name is needed. Read the three matching pieces in order around each triangle: leg, right angle, hypotenuse. The right angle sits between one leg and the hypotenuse only if you look at it the wrong way — properly, the equal angle ($90°$) is not included between the two sides you were given equal ($AB$/$DE$ and $AC$/$DF$; the leg-and-hypotenuse pair straddles the right angle rather than sandwiching it). So this is structurally a side-side-angle (SSA) match, and SSA is not a valid congruence test for triangles in general — two different triangles can share two sides and a non-included angle. What rescues the right-triangle case is Pythagoras: once the hypotenuse and one leg of a right triangle are fixed, the third side is pinned down uniquely ($\text{third leg} = \sqrt{\text{hyp}^2 - \text{leg}^2}$, always positive and single-valued), so no second, different triangle can satisfy the same data. That is the whole reason RHS gets listed as a fourth, additional criterion instead of being folded into SAS or ASA.

Spotting an RHS setup. Look for a figure that already hands you, or lets you deduce, a right angle in both triangles — a foot of a perpendicular, an altitude, a perpendicular bisector, a ladder against a wall — together with equal hypotenuses and one equal pair of legs.

RHS is stated without proof at this level, so instead of a derivation this section makes the reasoning behind it precise using Pythagoras, which students already know. The claim that needs justifying is: given a right angle, one leg, and the hypotenuse of a right triangle, no second, genuinely different right triangle can be built from the same three numbers. Consider a right triangle with legs $p$ and $q$ and hypotenuse $h$, where $p$ and $h$ are the two given, fixed lengths and $q$ is unknown. Pythagoras' theorem gives

where: $p$ is the known leg, $h$ is the known hypotenuse, and $q$ is the remaining leg to be found. Solving for the unknown leg, $q = \sqrt{h^2 - p^2}$. Since $h > p$ in any right triangle (the hypotenuse is always the longest side), $h^2 - p^2$ is a positive number, and a positive number has exactly one positive square root. So $q$ is fixed — uniquely determined by $p$ and $h$, with no ambiguity and no second possible value. That single-valuedness is precisely what SSA lacks in general triangles (there, the missing side can typically be completed in two different ways, giving two different triangles from the same SSA data). Because the right angle removes that ambiguity, two right triangles sharing a hypotenuse and one leg must also share the third leg, and hence be congruent by SSS reasoning underneath the surface — which is the real engine powering RHS.

$AD$ is the altitude from vertex $A$ of an isosceles triangle $ABC$ (with $AB = AC$) to the base $BC$, meeting it at $D$, so $AD \perp BC$. Prove $\triangle ABD \cong \triangle ACD$ using RHS, and hence show that $D$ is the midpoint of $BC$.

Step 1. Identify the right angles. Since $AD \perp BC$, both $\angle ADB$ and $\angle ADC$ equal $90°$:

So $\triangle ABD$ and $\triangle ACD$ are both right triangles, right-angled at $D$ — the first condition RHS needs.

Step 2. Identify the hypotenuses and match them. In $\triangle ABD$, the side opposite the right angle at $D$ is $AB$; in $\triangle ACD$, it is $AC$. Since the triangle is isosceles with $AB = AC$ given, the hypotenuses are equal:

Step 3. Identify the equal leg. $AD$ is a side of both triangles — the altitude itself, shared by both:

This is the second condition: one pair of corresponding legs equal.

Step 4. Apply RHS and conclude with CPCT. With $\angle ADB = \angle ADC = 90°$, hypotenuses $AB = AC$, and legs $AD = AD$, the RHS criterion gives $\triangle ABD \cong \triangle ACD$. By CPCT (corresponding parts of congruent triangles), $BD = CD$, so $D$ is the midpoint of $BC$. The altitude from the apex of an isosceles triangle bisects the base — confirmed here via RHS rather than the SAS route seen elsewhere.

Competitive stretch

Suppose instead you are told $AB = AC$ and $BD = CD$ (not that $AD \perp BC$), and asked to prove $AD \perp BC$. Here $\triangle ABD$ and $\triangle ACD$ share $AD$ and have $AB = AC$, $BD = CD$ — that is SSS, giving $\triangle ABD \cong \triangle ACD$ directly, and CPCT then gives $\angle ADB = \angle ADC$. Since these angles are supplementary (they lie on straight line $BC$) and equal, each must be $90°$, so $AD \perp BC$. This flips the logic of the worked example: SSS proves the perpendicularity that RHS had assumed, a reminder that the same triangle pair can be attacked by more than one criterion depending on what is given first.

• Invoking RHS without a stated or provable right angle in both triangles — "the figure looks square" is not evidence; the $90°$ must be justified.
• Mislabelling the hypotenuse — it is always the side opposite the right angle, never one of the two legs that form it.
• Reaching for RHS whenever a right angle appears, even when the two given equal sides are the two legs with the right angle included between them — that configuration is SAS, not RHS.
• Skipping the correct vertex correspondence before writing CPCT conclusions — RHS still requires matching $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$ correctly, or the "corresponding" parts read off afterward will be wrong.

Ask the tutor: If a figure has a right angle but you are unsure whether to use RHS or SAS, ask the AI tutor to check with you whether the right angle sits between the two given equal sides.