Areas of Parallelograms and Triangles
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
Sponsor Area
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.
Sponsor Area
Sponsor Area