Question

From the following figure, prove that : (i) angle ACD = angle CBE (ii) AD = CE​

Answer 1

Answer:

here's your answer

Step-by-step explanation:

AC=AB

angle ACB=angle ABC

180°-ACB=180°-ABC

angle ACD=angle CBE

in ∆ACD and ∆CBE

DC=CB

AC=CE

angle ACD=angle CBE

∆ACD is congruent to ∆CBE

SSA congruency

by CPCT, AC=CE

Answer 2

Answer: 1) IN TRI. ACD AND TRI. ACB

                      DC=CB-GIVEN

                      AC=AB-GIVEN

                      AC=AC-COMMON SIDE

THEREFORE: TRI.DAC IS CONGRUENT TO TRI. CAB by S-S-S TEST

                        therefore: angleDAC=angleBAC

                                          angleACD=angleABC........(i)

                     IN TRI. ABC and TRI. BCE

                         BE=AB-GIVEN

                         BC=BC-COMMON SIDE

                         AC=BE-GIVEN

 THEREFORE: TRI.ABC is congruent to TRI.CBE by S-S-S test

 THEREFORE :  angleABC=angleCBE.......(ii)

FROM 1 AND 2 WE KNOW THAT

angleACD=angleCBE......(iii)

2) IN TRI. ACD and TRI.CBE

      AC=BE-GIVEN

     DC=CB-GIVEN

    angleACD=angleCBE-FROM 3

THEREFORE.TRI.ACD is congruent to TRI.CBE by S-A-S test

therefore we can say that AD=CE

HENCE PROVED angleACD=angleCBE and AD=CE

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