(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 ∆OAP and ∆OBQ,
AP = BQ | Given
∠OAP = ∠OBQ | Each = 90°
∠AOP = ∠BOQ
| Vertically Opposite Angles
∴ ∆OAP ≅ ∆OBQ | AAS Axiom
∴ OA = OB | C.P.C.T.
and OP = OQ | C.P.C.T.
⇒ O is the mid-point of line segments AB and PQ