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In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$. If $CD = 7$ and $[\Delta ABD] = a\sqrt{5}$ , find $a$ .

What I Tried: Here is a picture:-

Let the perpendicular bisector of $BC$ pass through $BC$ at $E$ .
Then I first noticed that $\Delta BDE \cong \Delta CDE$ from $(SAS)$ congruency.
This gives the required information in the diagram, as well as we have $BD = 7$ .
Now :- $$\Delta ABD \sim \Delta ACB$$ $$\rightarrow \frac{AD}{AB} = \frac{7}{BC} = \frac{AB}{AC}$$ So let $AD = k$ , $AB = m$ , $BE = EC = n$ . We have :- $$\frac{k}{m} = \frac{7}{2n} = \frac{m}{(7+k)}$$

Anonymous
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  • You are correct, there is not enough information to solve the problem without doing things in terms of $x$. You can see this by constructing an arbitrary $\triangle BCD$ such that $BD = CD = 7$. From there, you can use the angles you found to construct the rest of the triangle. This will work for any $\triangle BCD$. By the way, it's actually $2$ equations in $4$ variables because $a = b$ and $b = c$ implies $a = c$. – Joshua Wang Nov 13 '20 at 17:34
  • Ok, so what should I do next? – Anonymous Nov 13 '20 at 17:51
  • To find the area in terms of $x$, I would drop a perpendicular from $D$ to $AB$ (foot $F$). Then, $DE = DF$. You can use trigonometry to find $AD$, from which you can find $[ABD]$. – Joshua Wang Nov 13 '20 at 17:54
  • Why is $DE = DF$? Can you explain more, or I am missing something? – Anonymous Nov 13 '20 at 18:47
  • Because $\angle DBE = \angle DEF, \angle DEB = \angle DFB = 90^{\circ}$, and $DB = DB$ we have a SAA congruence between $\triangle DEB$ and $\triangle DFB$ – Joshua Wang Nov 13 '20 at 18:49
  • What is $[\Delta ABD] $ area or perimeter? – Raffaele Nov 13 '20 at 19:35
  • The problem has no definite solution. Infinite different triangles like that can be built. – Raffaele Nov 13 '20 at 20:02
  • @Raffaele, $[\Delta ABD]$ is area. – Anonymous Nov 14 '20 at 05:15

2 Answers2

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Let $E=(0,0)$, $B=(-\alpha,0)$, $C=(\alpha,0)$, $D=(0,\beta)$. Then we have $$\alpha^2+\beta^2=7^2=49,\quad\tan \angle ACB=\frac{\beta}{\alpha}\in(0,\sqrt3).$$ Let $t=\tan \angle ACB$, then $$y_{AB}=\frac{2t}{1-t^2}(x+\alpha),\quad y_{AC}=-tx+\beta.$$ And we have $$A=\left(-\alpha\frac{1+t^2}{3-t^2},\alpha\frac{4t}{3-t^2}\right),\quad |AB|=2\alpha\frac{1+t^2}{3-t^2}.$$ Thus, the area \begin{align} S_{\triangle ABD} &= \frac12|AB|\cdot|DE|\\ &=\alpha\beta\frac{1+t^2}{3-t^2}\\ &=\frac{49t}{3-t^2}\in(0,\infty). \end{align} If you have some other conditions such that $\frac{t}{3-t^2}=\sqrt5$, then $a=49$. If not, $a$ can be any positive real number.

Aforest
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Note:- As others mentioned, this question was unsolvable before because there can be infinitely many such triangles. I also realized that is true because I missed a small detail that $AD = 9$. Sorry for the inconvenience. I also figured out how to solve this now.

I had figured out that :- $$\frac{k}{m} = \frac{7}{2n} = \frac{m}{(7+k)}$$ With the new information this becomes :- $$\frac{9}{m} = \frac{7}{2n} = \frac{m}{16}$$

From here, I get $m = 12$ , so I have the lengths of all the sides of $\Delta ABC$.
Now from Heron's Formula I have :- $$[\Delta ABD] = \sqrt{s(s-a)(s-b)(s-c)}$$ $$\rightarrow \sqrt{14 * 2 * 5 * 7}$$ $$\rightarrow \sqrt{980} = 14\sqrt{5}$$ So we have $a = 14$ as our solution.

Anonymous
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