(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Given: ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA.
To Prove: (i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Proof: (i) In ∆ABD and ∆BAC,
AD = BC | Given
AB = BA | Common
∠DAB = ∠CBA | Given
∴ ∆ABD ≅ ∠BAC | SAS Rule
(ii) ∵ ∆ABD ≅ ∆BAC | Proved in (i)
∴ BD = AC | C.P.C.T.
(iii) ∵ ∆ABD ≅ ∠BAC | Proved in (i)
∴ ∠ABD = ∠BAC. | C.P.C.T.
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