Question
In the given figure, prove that:
(i) $\triangle ACB \cong \triangle ECD$
(ii) $A B=E D$
(i) $\triangle ACB \cong \triangle ECD$
(ii) $A B=E D$
✓
Answer
(i) In Δ ACB and Δ ECD,
AC = CE ................(given)
∠ACB = ∠DCE ............(vertically opposite angles)
BC = CD .............(given)
∴ Δ ACB ≅ Δ ECD ..............(S.A.S. Axiom)
(ii) Hence AB = ED ................(c.p.c.t.)
Hence proved.
AC = CE ................(given)
∠ACB = ∠DCE ............(vertically opposite angles)
BC = CD .............(given)
∴ Δ ACB ≅ Δ ECD ..............(S.A.S. Axiom)
(ii) Hence AB = ED ................(c.p.c.t.)
Hence proved.
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