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This is the question:

$E$ is a point on the side $AB$ of a rectangle $ABCD$ such that $DE=6$ cm, $DA= 8cm$ and $DC=6cm$. If $CE$ extended meets the circumcircle of the rectangle at $F$, then what is the value of $BF$ ? (use $\sqrt{2}$=$1.414$).

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Here is how I tried to solve it

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In $\Delta BCD$, $\frac{CD}{BC} = \frac{DG}{BG}$ (Since CG is the Angle bisector of $\langle C$ )

From this, DG = $\frac{30}{7} cm$; BG= $\frac{40}{7} cm $

$\langle FCD $ = $\langle FAD$ = $\langle FBD$ = $45°$

Therefore, $\Delta CED$ ~ $\Delta AEF$ ; i.e. $\Delta AEF$ is a right isosceles triangle like $\Delta CED $

Hence, EF = AF = $\sqrt 2 cm $

In $\Delta CED $, CE = 6$\sqrt2$ cm

Now consider $\Delta CGD$ & $\Delta BGF $

$\Delta CGD$ ~ $\Delta BGF $

Assume the value of $GE = x$ $cm$

$\Delta GCD$ ~ $\Delta GBF$

$\frac {GC}{GB}$ = $\frac {CD}{BF}$ = $\frac {GD}{GF}$

$\frac {(6 \sqrt 2 -x)}{\frac{40}{7}}$ = $\frac {6}{BF}$ = $\frac {\frac{30}{7}}{(x+\sqrt 2)}$

$(x+\sqrt 2)(6 \sqrt 2 - x)$ = $\frac{30}{7}$ .$\frac{40}{7}$

My plan was to find the value of $x$ so that I could get the ratio and then evaluate $BF$. However, it looks like I am going wrong as I am getting complicated values for $x$. Could anyone please guide?

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  • $\begingroup$ Did you mean to say that E is on the side AD? $\endgroup$ – steven gregory Jan 2 '18 at 16:35
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By law of sines for $\Delta CFB$ we obtain: $$BF=AC\cdot\sin45^{\circ}=\sqrt{6^2+8^2}\sin45^{\circ}=5\sqrt2.$$

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  • $\begingroup$ I am sorry, but to which triangle are you applying the law of sines? $\endgroup$ – Fasal123 Jan 2 '18 at 16:30
  • $\begingroup$ In $\Delta CFB$ we have: $BF=2R_{\Delta CFB}\sin\measuredangle FCB$. $\endgroup$ – Michael Rozenberg Jan 2 '18 at 16:33
  • $\begingroup$ Thanks. I seem to have complicated it unnecessarily. $\endgroup$ – Fasal123 Jan 2 '18 at 16:42
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Jan 2 '18 at 16:43
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If $DA=8$ and $DE=6$, then $AE=2$.

Since $AC$ and $BD$ are diagonals of a $6 \times 8$ rectangle, $AC = BD = 10.$

$\triangle CDE$ is an isosceles right triangle. So $CE = 6 \sqrt 2.$

$\triangle CDE \sim \triangle AFE$. So $\triangle AFE$ is an isosceles right triangle. Since $AE=2$, then $EF = AF = \sqrt 2.$

$\triangle CAE \sim \triangle DFE \implies \dfrac{CA=10}{\color{red}{DF=5\sqrt 2}} = \dfrac{CE=6\sqrt 2}{DE=6} = \dfrac{AE=2}{FE=\sqrt 2}$

Since $BD$ is a diameter of the circle, $\angle BFD$ is a right angle. Hence $\triangle BFD$ is a right triangle. So $BF = \sqrt{BD^2 - DF^2} = \sqrt{10^2 - (5\sqrt 2)^2} = 5\sqrt 2$

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  • $\begingroup$ Thanks for the answer. Since your approach is similar to mine, could you please explain where I was wrong? $\endgroup$ – Fasal123 Jan 3 '18 at 13:50
  • $\begingroup$ You weren't wrong. You get $x \in \{ \dfrac{17}{7}\sqrt 2,\dfrac{18}{7}\sqrt 2\}$ and I can't see immediately how to choose one over the other. You should have just looked for something else. $\endgroup$ – steven gregory Jan 3 '18 at 16:00

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