(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
In ∆QPR and ∆PQS,
QR = PS Â Â Â | Given
∠RQP = ∠SPQ    | Given
PQ = PQ Â Â Â | Common
∴ ∆QPR ≅ ∆PQS    | SAS Axiom
∴ PR = QS    | C.P.C.T.
and    ∠QPR = ∠PQS.    | C.P.C.T.