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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

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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

Guest Feb 12, 2015

#1
+20024
+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
#1
+20024
+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
#2
+89876
+3

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

CPhill  Feb 12, 2015
#3
+93644
+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.

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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
+20024
+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