13 In the following diagram. ABC is a triangle in which is the midpoint of BC, and is a point on AC such that AE: EC = 2:1 (see bottom of page for HINT] 10A = 2.03 - and OC = $. find: (a) ( 0B C 131 @ DE (3) (W) DE 121 Анти Ag - 28-09 - =
(b) () Find BA + CA + 2B0 [2] (1) Hence deduce that BA + CA + 2BC is parallel to DE.
13 In the following diagram. ABC is a triangle in which is the midpoint of BC, and is a point on AC such that AE: EC = 2
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13 In the following diagram. ABC is a triangle in which is the midpoint of BC, and is a point on AC such that AE: EC = 2
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