RHS Congruence Rule (original) (raw)

Last Updated : 23 Jul, 2025

RHS Congruence Rule is also known as the HL (Hypotenuse-Leg) Congruence Theorem. It states the criteria for any two right-angle triangles to be congruent.

This rule states that if in two right triangles, the hypotenuse and one side of one triangle are equal to the hypotenuse and one corresponding side of the other triangle, then the triangles are congruent. In this article, we will discuss the criteria of congruence of right-angle triangles in detail including proof and examples.

Table of Content

• RHS Congruence Rule
• Proof
• Steps to apply RHS Congruence Rule
• RHS and SSS Congurence Rule
• Solved Examples

What is RHS Congruence Rule?

**RHS stands for Right Angle-Hypotenuse-Side.

RHS Congruence Rule states that in two right-angled triangles, if the length of the hypotenuse and one side of one triangle is equal to the length of the hypotenuse and corresponding side of the other triangle, then the two triangles are congruent

RHS Criterion of Congreunce

In symbols, if in two right triangles △ABC and △__DEF, we have:

• __AB_=_DE (The hypotenuses are congruent).
• ∠_B_=∠__E (Both triangles have a right angle).
• __BC_=_EF (One corresponding side of each triangle is congruent).

△ABC ≅ △DEF (by RHS)

Proof of RHS Congruence Rule

Let's consider two right angle ΔABC and ΔDEF,

Where,

• ∠B = 90° and ∠E = 90°,
• Hypotenuse is equal i.e. AC = DF, and
• One side is equal i.e. BC = EF.

This, can be illustrated as follows.

**To Proof: We need to prove that ΔABC and ΔDEF are congruent.

**Proof:

In right ΔABC, By Pythagoras theorem,

AC2 = AB2 + BC2

⇒ AB2 = AC2 - BC2 . . . (1)

In right ΔDEF, By Pythagoras theorem,

DF2 = DE2 + EF2

⇒ DE2 = DF2 - EF2 . . . (2)

From (1),

AB2 = AC2 - BC2

⇒ AB2 = DF2 - EF2 (∵AC = DF and BC= EF (given))

⇒ AB2 = DE2 (From (2))

⇒ AB = DE . . .(3)

In ΔABC and ΔDEF,

• AB = DE (From Equation3)
• BC= EF (Given)
• AC = DF (Given)

ΔABC ≅ ΔDEF (By SSS congruence rule) [Hence proved.]

The above proves the RHS Congruence Rule.

How to apply RHS Congruence Rule?

To check if you can apply the RHS Congruence Rule to prove whether triangles are congruent, check the given and the triangles.

• Firstly, both triangles must be right-angled triangles or right triangles. It means, the one angle in both triangles should be 90° or a right angle.
• Next, it should be given, or measurable, or you must be able verify or confirm that the hypotenuse of both triangles are equal and any one side of one triangle is equal to the corresponding side of the other triangle.

If all the three conditions above meet then you can apply the RHS Congruence Rule to prove them to be congruent to each other.

RHS and SSS Congurence Rule

Key differences between the RHS and SSS congruence rules are listed in the following table:

**Read More,

• **Triangles in Geometry
• **Congruence of Triangles
• **Angle Sum Property of a Triangle
• **Similar Triangles

Examples on RHS Congruence Rule

**Example 1: P is a point equidistant from two lines l and m intersecting at point A. Show that the line AP bisects the angle between them.

**Solution:

In the question we are given that lines l and m intersect each other at A.

Let PB ⊥ l, PC ⊥ m.

It is also given that PB = PC.

We need to show that ∠ PAB = ∠ PAC

Considering Δ PAB and Δ PAC,

PB = PC (Given)

∠PBA = ∠PCA = 90° (Given)

PA = PA (Common)

So, by RHS congruency rule,

Δ PAB ≅ Δ PAC

Therefore we prove, ∠ PAB = ∠ PAC (∵ CPCT)

**Example 2: State and prove whether given triangles in the following image are congruent or not.

**Solution:

In the given triangles, △ZXY and △PQR,

• XZ = PQ [side]
• YZ = PR [hypotenuse]
• ∠ZXY= ∠PQR=90° [right angle]

∴△ZXY≅△PQR, by RHS congruence criterion.

Hence proved.

**Question 3: In the given triangle, △ABD, if AC bisects side BD and CE=CF, prove that the area of triangles △BCE and △DCF are equal.

**Solution:

Two congruent triangles are always equal in area. SO, we need to prove that both the triangles are congruent for solving this question.

△BCE and △DCF are right triangles, in which,

• CB = CD (as AC bisects BD)
• CE = CF (given)
• ∠CEB=∠CFD=90°

∴ △ BCE ≅ △ DCF (by RHS congruence criterion)

Hence, △BCE and △DCF are equal in area. **[Hence Prooved]

Practice Problems on RHS Congruence Rule

**Problem 1: Triangle ABC has a right angle at B, where AB = 10, BC = 24, and AC = 26. Triangle DEF has a right angle at E, where DE = 10 and DF = 26. Are the two triangles congruent?

**Problem 2: Triangle PQR has a right angle at Q, where PQ = 15, QR = 20, and PR = 25. Triangle XYZ has a right angle at Y, where XY = 20 and XZ = 25. Are the two triangles congruent?

**Problem 3: Triangle LMN has a right angle at M, where LM = 9, MN = 12, and LN = 15. Triangle STU has a right angle at T, where ST = 9 and SU = 15. Are the two triangles congruent?

**Problem 4: Triangle JKL has a right angle at J, where JK = 8, JL = 17, and KL = 15. Triangle VWX has a right angle at V, where VW = 8 and VX = 15. Are the two triangles congruent?