MATHS [2 Mark Question Answer]

Proof:
In right Δ ABD and Δ ABC
Side AB = Side AB ...................(common)
Hypotenuse AD = Hypotenuse AC ...............(given)
∴ Δ ABD ≅ Δ ABC ................(R.H.S. Axiom)
Hence BD = BC .................(c.p.c.t.)
Hence proved.

Proof:
(i) In ∆ ABC and ∆ ADC
AC = AC ...........(common)
AB = DC ............(given)
BC = AD ...........(given)
∴ ∆ ABC ≅ ∆ ADC ................(S.S.S. Axiom)
(ii) Hence ∠B = ∠D ...........(c.p.c.t.)
Hence proved.

(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.