Given: In quadrilateral ACBD, AC = AD and AB bisects ∠A.
To Prove: ∆ABC ≅ ∆ABD.
Proof: In ∆ABC and ∆ABD,
AC = AD Â Â Â | Given
AB = AB Â Â Â | Common
∠CAB = ∠DAB
| ∵ AB bisects ∠A
∴ ∠ABC ≅ ∠ABD    | SAS Rule
∴ BC = BD    | C.P.C.T,
(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Line I is the bisector of an angle ∠A and B is any point on I. BP and BQ are perpendiculars from B to the arms of ∠A (see figure). Show that:
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE.
(i)    ∆AMC ≅ ∆BMD
(ii)    ∠DBC is a right angle
(iii)    ∆DBC ≅ ∆ACB