# RHS (Right Angle-Hypotenuse-Side) Congruence Theorem

**Theorem : ** Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle.

**Given: **

In Î”ABC and Î”DEF

âˆ B = âˆ E = 90Â°

AC = DF

BC = EF

**To Prove: ** Î”ABC â‰… Î”DEF

**Construction: ** Produce DE to M so that EM = AB, Join MF.

**RHS (Right Angle-Hypotenuse-Side) Congruence Theorem**

**Proof:**

In Î”ABC and Î”MEF,

AB = ME ... (by construction)

BC = EF ... (Given)

âˆ B = âˆ MEF ... (each 90Â°)

âˆ´ Î”ABC â‰… Î”MEF ... (SAS Cong. Axiom)

Hence, âˆ A = âˆ M ... (c.p.c.t.) ...(i)

AC = MF ... (c.p.c.t.)...(ii)

Also, AC = DF ... (Given)

âˆ´ DF = MF

âˆ´ âˆ D = âˆ M ... (âˆ s opposite to equal sides of Î”DFM) ...(iii)

From (1) and (3):

âˆ A = âˆ D ...(iv)

Now, In Î”ABC and Î”DEF,

âˆ A = âˆ D ... (From (4))

âˆ B = âˆ E ... (Given)

âˆ´ âˆ C = âˆ F ... (v)

Again, In Î”ABC and Î”DEF,

BC = EF ... (Given)

AC = DF ... (Given)

âˆ C = âˆ F ... (From (v))

âˆ´ Î”ABC â‰… Î”DEF ... (SAS Cong. Axiom)

**Remark: **If the three angles of one triangle is equal to the corresponding angles of the other triangle, the two triangles will not necessarily be congruent (they will be similar).