dear student let AB⇀=a⇀ and AD⇀=b⇀then BC⇀=AD⇀=b⇀ and AC⇀=AB⇀+AD⇀=a⇀+b⇀also let K be a point on AC, such that AK:AC=1:3or AK=13ACAK⇀=13(a⇀+b⇀) --- (1)again E being the midpoint of AB, we haveAE⇀=12a⇀Let M be the midpoint of DE such that DM:ME=2:1AM⇀=AD⇀+2AE⇀1+2=b⇀+a⇀3---- (2)From 1 and 2, we find thatAK⇀=13(a⇀+b⇀)=AM⇀and so we conclude that K and M coincide i e DE trisect AC and is trisected by AC regards

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dear student

l e t \overset{⇀}{A B} = \overset{⇀}{a} a n d \overset{⇀}{A D} = \overset{⇀}{b} t h e n \overset{⇀}{B C} = \overset{⇀}{A D} = \overset{⇀}{b} a n d \overset{⇀}{A C} = \overset{⇀}{A B} + \overset{⇀}{A D} = \overset{⇀}{a} + \overset{⇀}{b} a l s o l e t K b e a p o i n t o n A C, s u c h t h a t A K : A C = 1 : 3 o r A K = \frac{1}{3} A C \overset{⇀}{A K} = \frac{1}{3} (\overset{⇀}{a} + \overset{⇀}{b}) - - - (1) a g a i n E b e i n g t h e m i d p o i n t o f A B, w e h a v e \overset{⇀}{A E} = \frac{1}{2} \overset{⇀}{a} L e t M b e t h e m i d p o i n t o f D E s u c h t h a t D M : M E = 2 : 1 \overset{⇀}{A M} = \frac{\overset{⇀}{A D} + 2 \overset{⇀}{A E}}{1 + 2} = \frac{\overset{⇀}{b} + \overset{⇀}{a}}{3} - - - - (2) F r o m 1 a n d 2, w e f i n d t h a t \overset{⇀}{A K} = \frac{1}{3} (\overset{⇀}{a} + \overset{⇀}{b}) = \overset{⇀}{A M} a n d s o w e c o n c l u d e t h a t K a n d M c o i n c i d e i . e . D E t r i s e c t A C a n d i s t r i s e c t e d b y A C

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