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IN TRIANGLE ABC AND TRIANGLE EDC
AB = DE [GIVEN]
ANGLE C = ANGLE C [COMMON]
AND BC = DC GIVEN
THEREFORE TRIANGLE ABC CONGRUENT TO TRIANGLE EDC BY SAS CONGRUENCY RULE
AB = DE [GIVEN]
ANGLE C = ANGLE C [COMMON]
AND BC = DC GIVEN
THEREFORE TRIANGLE ABC CONGRUENT TO TRIANGLE EDC BY SAS CONGRUENCY RULE
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consider ∆abc and ∆edc,we get
ab=de(given)
bc=dc(given
angle c =angle c
by SSA
∆abc≈∆edc
ab=de(given)
bc=dc(given
angle c =angle c
by SSA
∆abc≈∆edc
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