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Equal Intercept Theorem

In the fig sh...

Question

In the fig shown $D E ∥ B C$ and $A D = 3 x - 2, A E = 5 x - 4, B D = 7 x - 5, C E = 5 x - 3$ then find the value of $x$

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Solution

REF.Image
In Given fig. DE||BC
now using mid point theorem,
$\frac{A D}{B D} = \frac{A E}{E C}$
Given AD=3x-2, AE=5x-4, BD= 7x-5 & CE=5x-3
$\frac{3 x - 2}{7 x - 5} = \frac{5 x - 4}{5 x - 3} \Rightarrow (3 x - 2) (5 x - 3) = (5 x - 4) (7 x - 5)$
$\Rightarrow 15 x^{2} - 19 x + 6 = 35 x^{2} - 53 x + 20$
$\Rightarrow 20 x^{2} - 34 x + 14 = 0 \Rightarrow 10 x^{2} - 17 x + 7 = 0$
$(x - 1) (10 x - 7) = 0$
Hence [x=1] & $[x = \frac{7}{10}]$
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Q. In the figure if

D E ∥ B C

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A E = 5 x - 4

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