If ∆ABC ≅ ∆DEF by SSS congruence rule then:

• AB = EF, BC = FD, CA = DE
• AB = FD, BC = DE, CA = EF
• AB = DE, BC = EF, CA = FD
• AB = DE, BC = EF, ∠C = ∠F

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.

 

AB is a line-segment. AX and BY are two equal line-segments drawn on opposite sides of line AB such that AX || BY. If AB and XY intersect each other at P. Prove that:
(i)    ∆APX ≅ ∆BPY
(ii)    AB and XY bisect each other at P.