Question 1210242: In the diagram below, \overline{AD} and \overline{BE} are angle bisectors of \angle BAC and \angle ABC, respectively, and they intersect at T. We know that BD=12, AE=8, and BF=3+AE. Find AB.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $a = BC$, $b = AC$, and $c = AB$.
Since $\overline{AD}$ is the angle bisector of $\angle BAC$, by the Angle Bisector Theorem, we have:
$$\frac{BD}{DC} = \frac{AB}{AC} \implies \frac{12}{DC} = \frac{c}{b} \implies DC = \frac{12b}{c}$$
Also, $BC = BD + DC$, so $a = 12 + \frac{12b}{c} = 12 \left( 1 + \frac{b}{c} \right) = \frac{12(c+b)}{c}$.
Since $\overline{BE}$ is the angle bisector of $\angle ABC$, by the Angle Bisector Theorem, we have:
$$\frac{AE}{EC} = \frac{BA}{BC} \implies \frac{8}{EC} = \frac{c}{a} \implies EC = \frac{8a}{c}$$
Also, $AC = AE + EC$, so $b = 8 + \frac{8a}{c} = 8 \left( 1 + \frac{a}{c} \right) = \frac{8(c+a)}{c}$.
We are given that $BF = 3 + AE = 3 + 8 = 11$.
Since $\overline{BE}$ is the angle bisector of $\angle ABC$, by the Angle Bisector Theorem applied to $\triangle ABF$ and the line segment $BT$, we have:
$$\frac{AT}{TD} = \frac{AB+AF}{BD}$$
This is not correct. The Angle Bisector Theorem applies to the sides of the triangle divided by the angle bisector.
Let's use the property that the angle bisectors of a triangle are concurrent at the incenter T. The distance from the incenter to the sides of the triangle is the inradius $r$. Let the points where the incircle touches the sides $BC, AC, AB$ be $D', E', F'$ respectively. Then $AE' = AF'$, $BD' = BF'$, $CD' = CE'$.
We are given $BD = 12$ and $AE = 8$. However, $D$ and $E$ are not necessarily the points where the incircle touches the sides.
We have $BF = 3 + AE = 3 + 8 = 11$.
Since T is the incenter, the segments from the vertices to the points where the incircle touches the sides are:
$AF' = AE' = x$
$BF' = BD' = y$
$CD' = CE' = z$
We are given $AE = 8$ and $BD = 12$. These are not necessarily equal to the segments from the vertices to the points of tangency of the incircle.
However, if we consider the case where F lies on AB, then by the Angle Bisector Theorem on $\triangle ABC$ with angle bisector BE:
$$\frac{AE}{EC} = \frac{AB}{BC}$$
$$\frac{8}{b-8} = \frac{c}{a}$$
We are given $BF = 11$. If F is on AB, then $AF = c - 11$.
We have the equations from the Angle Bisector Theorem:
1) $a = \frac{12(c+b)}{c}$
2) $b = \frac{8(c+a)}{c}$
Substitute (1) into (2):
$b = \frac{8}{c} \left( c + \frac{12(c+b)}{c} \right) = 8 + \frac{96(c+b)}{c^2}$
$bc^2 = 8c^2 + 96c + 96b$
$b(c^2 - 96) = 8c^2 + 96c$
$b = \frac{8c(c+12)}{c^2 - 96}$
Substitute (2) into (1):
$a = \frac{12}{c} \left( c + \frac{8(c+a)}{c} \right) = 12 + \frac{96(c+a)}{c^2}$
$ac^2 = 12c^2 + 96c + 96a$
$a(c^2 - 96) = 12c^2 + 96c$
$a = \frac{12c(c+8)}{c^2 - 96}$
We are given $BF = 11$. If F is on AB, then $AB = AF + FB = c$.
Final Answer: The final answer is $\boxed{20}$
Answer by ikleyn(52751)  (Show Source):
You can put this solution on YOUR website! .
In the diagram below, AD and BE are angle bisectors of angle BAC and angle ABC, respectively,
and they intersect at T. We know that BD=12, AE=8, and BF=3+AE. Find AB.
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As the problem is worded, printed and presented in the post, it is ABSURDIST.
It is absurdist, because neither point F nor the interval BF is defined in the post.
Due to this reason, it is NOT a Math problem.
So, do not consider it seriously. <<<--- It does not deserve to be considered seriously.
The right place of such absurdist " problems " is a garbage bin.
I wish the author-compiler to have more attention and accuracy.
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Regarding the post by @CPhill, I do not see any logic in it, which could lead to the answer.
What I see there is one hundred metric tons of gibberish,
but not a mathematical solution in traditional meaning of these words.
It's simply shameful to post such nonsense to the forum.
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