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Question

If A, B, C, D, E are five points in a plane such that AB+AE+BC+DC+ED=k AC, then the value of k is _______________.

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Solution

Given: AB+AE+BC+DC+ED=k AC


AB+AE+BC+DC+ED=k ACAB+BC+DC+AE+ED=k ACAC+DC+AD=k ACAC+AD+DC=k ACAC+AC=k AC2AC=k AC2=k


Hence, the value of k is 2.

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