Question

In a trapezium $ABCD$, $AB$ is parallel to $DC$ and $DC = 2AB$. $EF$ is parallel to $AB$, where $E$ and $F$ lie on $BC$ and $AD$ respectively such that $\dfrac{BE}{EC} = \dfrac{4}{3}$. Diagonal $DB$ intersects $EF$ at $G$. Prove that:

$5AB^2 = 5AC^2 + BC^2$

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Answer 1

In a trapezium ABCD, AB||DC and DC = 2AB

Also, $\sf\frac{BE}{EC}$=[tex}\sf\frac{4}{3}[/tex]

In trapezium ABCD, EF||AB||CD

$\therefore$$\sf\frac{ AF }{FD} = \frac{BE}{EC} = \frac{4}{3}$

In $\bf ΔBGE \: and \: ΔBDC,$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: ∠B = ∠B \: (common)$

$\therefore$$\sf ΔBGE \:≈ \: ΔBDC \: [AA \: similarly ]$

$\sf \frac{EG}{CD} = \frac{BE}{BC} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ----eqtn(i)$

$\sf \frac{be}{ec} = \frac{4}{3}$

$\sf \frac{EC}{be} = \frac{3}{4}$

$\sf \frac{EC}{EC} + 1 = \frac{3}{4} + 1$

$\sf \frac{EC + BE}{BE} = \frac{7}{4}$

$\sf \frac{BE}{BC} = \frac{4}{7}$

from eqtn (I),

$\sf \frac{EG}{CD} = \frac{4}{7}$

$\sf EG= \frac{4}{7} CD \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ----eqtn(ii)$

similarly, ΔDGF ~ ΔDBA

$\sf \frac{DE}{DA} = \frac{FG}{AB}$

$\sf \frac{FG}{AB} = \frac{3}{7}$

$\sf FG = \frac{3}{7} AB \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - - - eqtn(iii)$

$\boxed{\sf \frac{AF}{AD} = \frac{4}{7} = \frac{BE}{BC} \frac{EC}{BC} = \frac{3}{7} = \frac{DE}{DA} }$

Adding eqtn (II) and (III)

$\sf EG+FG = \frac{4}{7}CD+ \frac{3}{7}AB$

$\sf EF=\frac{4}{7}×(2AB) + \frac{3}{(AB)}$

$\sf = \frac{8}{7}AB+ \frac{3}{7}AB$

$\sf = \frac{11}{7}AB$

7EF= 11AB Hence, proved

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Hope you will understand.