ABC is a equilateral triangle. If #|BD|=|CF|# and #|CE|=6#, how do you find #|AD|=x#?

ABC is a equilateral triangle. If #|BD|=|CF|# and #|CE|=6#, how do you find #|AD|=x#?
enter image source here

2 Answers
Sep 12, 2017

#AD=x=12# units

Explanation:

enter image source here
Draw a line #DG#, parallel to #AC#, as shown in the figure.
#=> DeltaDBG# is an equilateral triangle.
Let #DB=y, => DG=BG=DB=y#,
Let #angleDEG=w, => angleFEC=w#,
#angleEGD=180-60=120^@#
#angleECF=180-60=120^@#
As #angleEGD=angleECF, angleDEG=angleFEC, and CF=DG#,
#=> DeltaEGD and DeltaECF# are congruent.
#=> EG=EC=6#

Given that #DeltaABC# is an equilateral triangle,
#=> AB=BC#,
#=> AD+DB=BG+GE+EC#,
#=> x+y=y+6+6#,
#=> x=12#

Hence, #AD=x=12# units

Sol. 2)

enter image source here
Draw a line #DH#, parallel to #BC#, as shown in the figure,
#=> DeltaADH# is an equilateral triangle,
#=> DH=AD=x#,
#=> HC=DB=y#,
#=> DeltaDHF and DeltaECF# are similar
#=> (DH)/(HF)=(EC)/(CF)#
#=> x/(2y)=6/y#,
#=> x=12# units

enter image source here

Sep 13, 2017

# x=12.#

Explanation:

Let us solve the Problem with the help of Trigonometry.

We want to apply the Sine-Rule to #DeltaDBE, and DeltaCEF.#

Hence, we need to find, from #DeltaDBE, #

#"sides "DB, BE, /_DEB, and, /_BDE.#

Similarly, in #DeltaCEF,# we will require,

#"side "CF, "/_EFC, and, /_CEF.#

Suppose that, #BC=a, and, /_EFC=theta.#

#DeltaABC# is equilateral.

#:. AB=BC=CA=a, and, /_B=60, /_ECF=120.#

In #DeltaCEF, /_EFC+/_ECF+ /_CEF=180.#

#:. theta+120+/_CEF=180 rArr /_CEF=60-theta=/_DEB.#

Also, #BD=AB-AD=a-x, &, as CF=BD=a-x.#

Similarly, #BE=BC-EC=a-6.#

In#DeltaCEF, EC=6, CF=a-x, /_EFC=theta, /_CEF=60-theta.#

#:. (CE)/(sin/_EFC)=(CF)/(sin/_CEF).......[because," the sine-rule]."#

#rArr 6/sintheta=(a-x)/sin(60-theta)....................(1).#

Next, in #DeltaDBC, DB=a-x, BE=a-6, /_DEB=60-theta, &#

#/_BDE=180-(/_DBE+/_DEB)=180-(60+60-theta)=60+theta.#

#:.," by the sine-rule, "(DB)/(sin/_DEB)=(BE)/(sin/_BDE).#

#rArr (a-x)/sin(60-theta)=(a-6)/sin(60+theta)..............(2).#

#(1), &, (2) rArr 6/sintheta=(a-6)/sin(60+theta), or,#

#sin(60+theta)/sintheta=(a-6)/6.#

#:. (sin60costheta+cos60sintheta)/sintheta=a/6-6/6.#

#:.sqrt3/2*costheta/sintheta+1/2*sintheta/sintheta=a/6-1.#

#:. sqrt3/2*cottheta+1/2+1=a/6, or, a=3sqrt3cottheta+9....(3).#

Now, from #(1), 6/sintheta=(a-x)/sin(60-theta).#

#:. sin(60-theta)/sintheta=(a-x)/6.#

#:. (sin60costheta-cos60sintheta)/sintheta=(a-x)/6.#

#:. sqrt3/2*costheta/sintheta-1/2*sintheta/sintheta=(a-x)/6.#

#:. 6(sqrt3/2*cottheta-1/2)=(a-x).#

#:. 3sqrt3cottheta-3-a=-x.#

Finally, by #(3),# then,

#x=a+3-3sqrt3cottheta=(3sqrt3cottheta+9)+3-3sqrt3cottheta.#

# rArr x=12,# as Respected CW has already obtained!

Enjoy Maths.!