bisector inside equilateral triangle

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bisector inside equilateral triangle

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#1 h Aug 9, 2012, 1:35 PM • 1 Y

Y by Adventure10

Let $ABC$ be an equilateral triangle, take line $t$ such that $t\parallel BC$ and $t$ passes through $A$ .
Let point $D$ be on side $AC$ , the bisector of angle $ABD$ intersects line $t$ in point $E$ .
Prove that $BD = CD + AE$ .

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#2 h Aug 9, 2012, 2:07 PM • 2 Y

Y by Adventure10, Mango247

mikeshadow wrote:

Let $ABC$ be an equilateral triangle, take line $t$ such that $t\parallel BC$ and $t$ passes through $A$ .
Let point $D$ be on side $AC$ , the bisector of angle $ABD$ intersects line $t$ in point $E$ .
Prove that $BD = CD + AE$ .

Let's rotate the triangle $\triangle{BDC}$ around the point $B$ at an angle of $60^0$ counterclockwise to the position $BD'A$ . Since $\angle{BCD}=60^0$ then $D' \in t$ . Notice that $\angle{D'BE}=\angle{CBE}=\angle{D'EB}$ . So $CD+AE=D'A+AE=D'E=D'B=DB$ , as desired.

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#3 h Aug 12, 2012, 9:46 AM • 2 Y

Y by Adventure10, Mango247

Take $M\in\overline{BD}|MD=CD$ , so we need to prove $BM=AE$ . Now let's take $N\in BE| AN=AE$ , then by angle chase we can easily prove $\triangle BCM\cong\triangle ABN$ by a.s.a. rule, hence $BM=AN=AE$ , done.

Best regards,
sunken rock

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#4 h Aug 12, 2012, 1:24 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ have side length $1$ . Let $\angle ABE = \angle DBE = x$ .
Note that $\angle DBC = 60^\circ - 2x$ , $\angle DCB = 60^\circ$ , $\angle BDC = 60^\circ + 2x$ and $\angle AEB = \angle EBC = 60^\circ - x$ .

The Sine Rule in $BDC$ gives $BD=\frac{\sqrt{3}}{2\sin(60^\circ + 2x)}$ and
$CD = \frac{\sin(60^\circ-2x)}{\sin(60^\circ+2x)}$ .
The Sine Rule in $AEB$ gives $AE = \frac{\sin x}{\sin(60^\circ - x)}$ .
$2\sin(60^\circ - x)\cos(60^\circ - x) = \sin(120^\circ - 2x) = \sin(60^\circ + 2x)$ so
$AE = \frac{2\sinh\cos(60^\circ - x)}{\sin(60^\circ + 2x}$

Hence we must prove that $\frac{\sqrt{3}}{2} = \sin(60^\circ - 2x) + 2\sinh\cos(60^\circ - x)$ , which holds because
$\sin(60^\circ - 2x) + 2\sin x\cos(60^\circ - x) = \frac{\sqrt{3}}{2}(1-2(\sin x)^2) - \frac{1}{2}2\sin x\cos x + 2\sin x(\frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x) = \frac{\sqrt{3}}{2}$ .

Done!

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#5 h Aug 15, 2012, 8:27 AM • 2 Y

Y by Adventure10, Mango247

We choose a point $E'$ on $(AC$ such that $[CE']\equiv [AE]$ . Now we see that:
$[BC]\equiv [AB]$ and $\angle BCE' \equiv \angle BAE$ $\Longrightarrow$ $\triangle BCE' \equiv \triangle BAE$ .
But $\angle DE'B = \angle AEB = \angle EBC = \angle EBD + \angle DBC = \angle DBC + \angle CBE'=$
$= \angle DBE'$ $\Longrightarrow$ $\triangle DBE'$ is isoscel.
$BD = DC + CE'=DC+AE$

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#6 h Apr 4, 2016, 5:07 PM • 2 Y

Y by Adventure10, Mango247

Let F be on t so A is between E and F and AF=DC
It results that FAB and DCB are congruent by the s.a.s case so BD=BF and m(<ABF)=m(<DCB)
so m(<FBE)=m(<EBC)
t is paralell to BC so m(<FEB)=m(<EBC)=m(<FBE) so the triangle FEB is isoscel with FE = FB <=> DC+AE=BD (QED)

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#7 h Apr 5, 2016, 3:09 PM • 2 Y

Y by Adventure10, Mango247

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#9 h Apr 12, 2023, 11:22 PM

Y by

thugzmath10 wrote:

Solution

What is point F?

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