+0  
 
0
280
4
avatar

In the figure:http://i57.tinypic.com/29ggy1u.jpg,D and E are the points lying on AC and AB respetively such that BE:AE =AD:DC=2:1.BD andCE intersects at K .Find EK:KC

Guest Feb 12, 2015

Best Answer 

 #1
avatar+19207 
+15

In the figure:http://i57.tinypic.com/29ggy1u.jpg,D and E are the points lying on AC and AB respetively such that BE:AE =AD:DC=2:1.BD andCE intersects at K .

Find EK:KC ?

$$\\\small{\text{We set the angles:}}\\
\small{\text{angle ABD $= B$ }}\\
\small{\text{angle ADB $ = D $ }}\\
\small{\text{angle BKE = angle BKC $ = K $}}\\\\
\small{\text{We set the line segments:}}\\
\small{\text{line segment BE $= b$ }}\\
\small{\text{line segment AE $= a$ }}\\
\small{\text{line segment CD $= c$ }}\\
\small{\text{line segment DA $= d$ }}\\
\small{\text{line segment EK $= p$ }}\\
\small{\text{line segment KC $= q$ }}\\$$

I.

$$\dfrac{\sin{(B)}} {d} = \dfrac{\sin{(D)}} {a+b}\quad \Rightarrow \quad
\dfrac{\sin{(B)}}{\sin{(D)}}= \dfrac{d}{a+b}$$

II.

$$\small{\text{$\sin{(180-D)} = \sin{D}$}}\\\\
\dfrac{\sin{(D)}} {q} = \dfrac{\sin{(K)}} {c}\quad \Rightarrow \quad
q= c \cdot \dfrac{\sin{(D)}}{\sin{(K)}}$$

III.

$$\dfrac{\sin{(K)}} {b} = \dfrac{\sin{(B)}} {p}\quad \Rightarrow \quad
p= b \cdot \dfrac{\sin{(B)}}{\sin{(K)}}$$

IV.

$$\dfrac{p}{q}
=
\dfrac{
b \cdot \dfrac {\sin{(B)}} {\sin{(K)}}
}
{
c \cdot \dfrac{\sin{(D)}}{\sin{(K)}} }
}
=\dfrac {b} {c} \cdot \dfrac{\sin{(B)}}{\sin{(D)}} }
=\dfrac {b} {c} \cdot \left(\dfrac{d}{a+b} \right) \\\\
=
\dfrac {1} { \dfrac{c}{d} }\cdot \left(\dfrac {1}{ \dfrac{a+b}{b} } \right)
=
\dfrac {1} { \dfrac{c}{d} }\cdot \left(\dfrac {1}{ \dfrac{a}{b}+\dfrac{b}{b}
} \right)
=
\dfrac {1} { \dfrac{c}{d} }\cdot \left(\dfrac {1}{ \dfrac{a}{b}+1
} \right)\\\\\\
\dfrac{c}{d} = \dfrac{a}{b} = \dfrac{1}{2} \\\\
\dfrac{p}{q}
=2\cdot \left(\dfrac {1}{ \dfrac{1}{2}+1 } \right)
= 2\cdot \left(\dfrac {1}{ \dfrac{3}{2} }\right)
=2\cdot \dfrac{2}{3} \\\\
\boxed{\dfrac{p}{q} = \dfrac{4}{3}}$$

EK:KC = 4 : 3

heureka  Feb 12, 2015
Sort: 

4+0 Answers

 #1
avatar+19207 
+15
Best Answer

In the figure:http://i57.tinypic.com/29ggy1u.jpg,D and E are the points lying on AC and AB respetively such that BE:AE =AD:DC=2:1.BD andCE intersects at K .

Find EK:KC ?

$$\\\small{\text{We set the angles:}}\\
\small{\text{angle ABD $= B$ }}\\
\small{\text{angle ADB $ = D $ }}\\
\small{\text{angle BKE = angle BKC $ = K $}}\\\\
\small{\text{We set the line segments:}}\\
\small{\text{line segment BE $= b$ }}\\
\small{\text{line segment AE $= a$ }}\\
\small{\text{line segment CD $= c$ }}\\
\small{\text{line segment DA $= d$ }}\\
\small{\text{line segment EK $= p$ }}\\
\small{\text{line segment KC $= q$ }}\\$$

I.

$$\dfrac{\sin{(B)}} {d} = \dfrac{\sin{(D)}} {a+b}\quad \Rightarrow \quad
\dfrac{\sin{(B)}}{\sin{(D)}}= \dfrac{d}{a+b}$$

II.

$$\small{\text{$\sin{(180-D)} = \sin{D}$}}\\\\
\dfrac{\sin{(D)}} {q} = \dfrac{\sin{(K)}} {c}\quad \Rightarrow \quad
q= c \cdot \dfrac{\sin{(D)}}{\sin{(K)}}$$

III.

$$\dfrac{\sin{(K)}} {b} = \dfrac{\sin{(B)}} {p}\quad \Rightarrow \quad
p= b \cdot \dfrac{\sin{(B)}}{\sin{(K)}}$$

IV.

$$\dfrac{p}{q}
=
\dfrac{
b \cdot \dfrac {\sin{(B)}} {\sin{(K)}}
}
{
c \cdot \dfrac{\sin{(D)}}{\sin{(K)}} }
}
=\dfrac {b} {c} \cdot \dfrac{\sin{(B)}}{\sin{(D)}} }
=\dfrac {b} {c} \cdot \left(\dfrac{d}{a+b} \right) \\\\
=
\dfrac {1} { \dfrac{c}{d} }\cdot \left(\dfrac {1}{ \dfrac{a+b}{b} } \right)
=
\dfrac {1} { \dfrac{c}{d} }\cdot \left(\dfrac {1}{ \dfrac{a}{b}+\dfrac{b}{b}
} \right)
=
\dfrac {1} { \dfrac{c}{d} }\cdot \left(\dfrac {1}{ \dfrac{a}{b}+1
} \right)\\\\\\
\dfrac{c}{d} = \dfrac{a}{b} = \dfrac{1}{2} \\\\
\dfrac{p}{q}
=2\cdot \left(\dfrac {1}{ \dfrac{1}{2}+1 } \right)
= 2\cdot \left(\dfrac {1}{ \dfrac{3}{2} }\right)
=2\cdot \dfrac{2}{3} \\\\
\boxed{\dfrac{p}{q} = \dfrac{4}{3}}$$

EK:KC = 4 : 3

heureka  Feb 12, 2015
 #2
avatar+85727 
+3

I haven't had time to digest this one, heureka.....but it looks pretty impressive!!

CPhill  Feb 12, 2015
 #3
avatar+92217 
+8

That really does look VERY impressive Heureka.

Hopefully I will have time to examine it properly later, because I don't have it now.  

------------------------------------------------------------------

 

Hi anon,  Please read this post.

How to upload an Image.

http://web2.0calc.com/questions/how-to-upload-an-image-nbsp_1

------------------------------------------------------------------

 

I have put the question and answer together for other peopl's benefit :))

 

In the figure:http://i57.tinypic.com/29ggy1u.jpg,D and E are the points lying on AC and AB respetively such that BE:AE =AD:DC=2:1.BD andCE intersects at K .Find EK:KC

 

Here is the pic

Melody  Feb 13, 2015
 #4
avatar+19207 
+10

$$\dfrac{BK}{KD}
=\dfrac {1} { \dfrac{a}{b} }\cdot \left(1+\dfrac{d}{c} \right) \\\\\\
\dfrac{a}{b} = \dfrac{1}{2} \qquad \dfrac{d}{c} = \dfrac{2}{1} \\\\
\dfrac{BK}{KD} =
2\cdot \left( 1+2 \right) = 6 \\\\
\boxed{\dfrac{BK}{KD} = \dfrac{6}{1}}$$

heureka  Feb 13, 2015

35 Online Users

avatar
avatar
avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details