Right Angle Hypotenuse Side Rhs Congruence Theorem

Right-Angle-Hypotenuse-Side (RHS) Congruence Theorem: Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle. Given: Two right triangles, Triangle ABC and Triangle DEF in which ∠B = 90o and ∠E = 90o AC=DF and BC=EF. To prove: Triangle ABC ≅ Triangle DEF Construction: Produce DE to G so that EG=AB. Join GF.

Proof: In Triangle ABC and Triangle GEF , we have AB=GE

(By construction)

∠B = ∠FEG = 90o and BC=EF

(Given)

Therefore, By SAS criterion of congruence

Triangle ABC ≅ Triangle GEF

So, ∠A = ∠G

… (1)

AC = GF

… (2) (cpctc)

Further, AC=GF From (2) and AC=DF (Given) Therefore DF=GF So, ∠D = ∠G

… (3) (Angles opposite to equal sides are equal)

From (1) and (3), we get

∠A = ∠D

… (4)

Thus, in Triangle ABC and Triangle DEF ,

∠A = ∠D

[From (4)]

∠B = ∠E

(Given)

So, ∠A + ∠B = ∠D + ∠E or 180o − ∠C = 180o − ∠F ( Since ∠A + ∠B + ∠C = 180o and ∠D + ∠E + ∠F = 180o , ∠A + ∠B = 180o − ∠C and ∠A + ∠B = 180o − ∠F ) So, ∠C = ∠F

… (5)

Now, in Triangle ABC and Triangle DEF , we have BC=EF

(Given)

∠C = ∠ F

[From (5)]

And AC=DF So, by SAS criterion of congruence Triangle ABC ≅ Triangle DEF

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