Congruence of Triangles By RHS Criteria


 
 
Concept Explanation
 

Congruence of Triangles by RHS Criteria

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 ABC and DEF in which large angle B=angle E=90^{circ},AC=DF,AB=DE

To Prove:  large Delta ABCcong Delta DEF

Construction: Produce FE to G so that GE = BC. Join GD.

Proof: In large Delta sABC;and; DEG, we have

       BC = GE                                                [By construction]

    large angle B=angle DEG=90^{circ}

and,   AB = DE                                                [Given]

So, by SAS criterion of congruence, we have

  large Delta ABCcong Delta DEG                               ....(i)

large Rightarrow ;;;angle C=angle G     and , AC = DG            [ C.P.C.T.]  ...(ii)

and,    AC = DF                              [Given]

Using (i) and (2) we have

      DF = DG 

large Rightarrow ;;angle F=angle G       [Angles opposite to equal sides in large Delta DGF are equal] ...(iii)

From (ii) and (iii), we get

       large angle C=angle F                                     .....(iv)

Thus, in large Delta sABC;and; DEF , we have

    large angle C=angle F                                     [From (iv)]...............(1)

    large angle B=angle E                                         [Given]..........................(2)

Adding equations (1) and (2), we have

large Rightarrow ;;angle C+angle B=angle F+angle E

large Rightarrow ;;180^{circ}-angle A=180^{circ}-angle D      [large because angle A+angle B+angle C=180^{circ} and large because angle D+angle E+angle F=180^{circ}]

angle D-angle A=180^{0}-180^{0}

Rightarrow angle D-angle A=0

large Rightarrow ;;angle A=angle D                                                 ....(v)

Now, in large Delta sABC;and ;DEF , we have

        AB = DE                               [ Given ]

      large angle A=angle D                      [ From (v)]

  and,  AC = DF                            [ Given ]

So, by SAS criterion of congruence, we have

     large Delta ABCcong Delta DEF

Hence,  large Delta ABCcong Delta DEF

Illustration: If large Delta ABC is an isosceles triangle such that AB = AC and AD is an altitude from A on BC. Prove that (i) large angle B=angle C  (ii) AD bisects BC   (iii) AD bisects large angle A.

Solution:  In right triangles ADB and ADC, we have

             Hyp. AB = Hyp. AC                [Given]

                 AD = AD                    [common side]

So, by RHS criterion of congruence, we have

           large Delta ABDcong Delta ACD

large Rightarrow ;;angle B=angle C, BD=DC;and;angle BAD=angle CAD    [ C.P.C.T.]

large Rightarrow ;;angle B=angle C,AD;bisects;BC ;and;angle A

 

Sample Questions
(More Questions for each concept available in Login)
Question : 1

In the given figure, find PM

Right Option : C
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Question : 2

If DC = SR, AC = PR, then bigtriangleup ADCcong bigtriangleup PSR by which criteria?

Right Option : B
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Explanation
Question : 3

In the figure shown above,AB = AC and D is the mid-point of BC. Then  bigtriangleup ABDcong bigtriangleup ACD by which criteria?

Right Option : A
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Explanation
 
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