Question
In the given figure, prove that:
(i) $\triangle ABC \cong \triangle DCB$
(ii) $AC = DB$

Answer

Proof:
In Δ ABC and Δ DCB,
CB = CB ...........(common)
∠ABC = ∠BCD ..........(each 90°)
and AB = CD .............(given)
(i) ∴ Δ ABC ≅ Δ DCB ..............(S.A.S. Axiom)
(ii) Hence AC = DB ............(c.p.c.t.)
Hence proved.

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