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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 ΔABC and ΔDEF
B = E = 90°
AC = DF
BC = EF


To Prove: ΔABC ΔDEF
Construction: Produce DE to M so that EM = AB, Join MF.
 

RHS (Right Angle-Hypotenuse-Side) Congruence Theorem


 

Proof:
In ΔABC and ΔMEF,
AB = ME ... (by construction)
BC = EF ... (Given)

B = MEF ... (each 90°)
ΔABC ≅ ΔMEF ... (SAS Cong. Axiom)
Hence,
A = M ... (c.p.c.t.) ...(i)
AC = MF ... (c.p.c.t.)...(ii)
Also, AC = DF ... (Given)
DF = MF
D = M ... (s opposite to equal sides of ΔDFM) ...(iii)
From (1) and (3):

A = D ...(iv)
Now, In
ΔABC and ΔDEF,
A = D ... (From (4))
B = E ... (Given)
C = F ... (v)
Again, In
ΔABC and ΔDEF,
BC = EF ... (Given)
AC = DF ... (Given)

C = F ... (From (v))
ΔABC ΔDEF ... (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).

 





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