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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

 Feb 12, 2015

Best Answer 

 #1
avatar+21860 
+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

 Feb 12, 2015
 #1
avatar+21860 
+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+98196 
+3

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

 Feb 12, 2015
 #3
avatar+99356 
+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

 Feb 13, 2015
 #4
avatar+21860 
+10

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

.
 Feb 13, 2015

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