In triangle ABC, BD is a median. CF intersects BD at E so that BE = ED. Point F is on AB. Then, if BF = 5. What is the value of AB?

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In triangle ABC, BD is a median. CF intersects BD at E so that BE = ED. Point F is on AB. Then, if BF = 5. What is the value of AB?

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Answer 1 · 2018-05-31T03:23:52

$A B = 15$ $AB=15$ units

Explanation:

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Let $| A B C |$ $|ABC|$ denote area of $A B C$ $ABC$
let $| A B C | = 12 a$ $|ABC|=12a$ ,
as $A D = D C, \Rightarrow | B D C | = | B D A | = \frac{12 a}{2} = 6 a$ $AD=DC, => |BDC|=|BDA|=(12a)/2=6a$
as $B E = E D, \Rightarrow | B E C | = | D E C | = \frac{6 a}{2} = 3 a$ $BE=ED, => |BEC|=|DEC|=(6a)/2=3a$ ,
let $| B F E | = x, \Rightarrow | D F E | = x$ $|BFE|=x, => |DFE|=x$ ,
$\Rightarrow | A F D | = | B D A | - | B D F | = 6 a - 2 x$ $=> |AFD|=|BDA|-|BDF|=6a-2x$
As $A D = D C, \Rightarrow | A F D | = | C F D |$ $AD=DC, => |AFD|=|CFD|$ ,
$\Rightarrow 6 a - 2 x = 3 a + x$ $=> 6a-2x=3a+x$
$\Rightarrow x = a$ $=> color(red)(x=a)$
$\Rightarrow | C F B | = 3 a + x = 4 a$ $=> |CFB|=3a+x=4a$
$\Rightarrow A B : F B = | A B C | : | C F B | = 12 a : 4 a = 3 : 1$ $=> AB:FB=|ABC|:|CFB|=12a:4a=3:1$
$\Rightarrow A B = 3 \cdot F B = 3 \times 5 = 15$ $=> AB=3*FB=3xx5=15$ units

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In triangle ABC, BD is a median. CF intersects BD at E so that BE = ED. Point F is on AB. Then, if BF = 5. What is the value of AB?

1 Answer

Explanation:

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