The triangle ABC has sides AB = 137, AC = 241 an BC = 200. There is a point D, on BC, such that both incircles of triangles ABD and ACD touch AD at the same point E. What is the length of CD?
A. 152
B. 174
C. 179
D. 183
E. 197
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touch AD at the same point E
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- sanju09
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- anshumishra
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Please refer to the diagram below :sanju09 wrote:The triangle ABC has sides AB = 137, AC = 241 an BC = 200. There is a point D, on BC, such that both incircles of triangles ABD and ACD touch AD at the same point E. What is the length of CD?
A. 152
B. 174
C. 179
D. 183
E. 197
[spoiler]https://gmatmaths.com/[/spoiler]
This is based on the theory : Two tangents can always be drawn to a circle from any point outside the circle, and these tangents are equal in length
x+y = 137 --- 1
x+z = 241 ---- 2
y+z+2w = 200 --- 3
Need to find : CD = CF"+DF" = w+z = ?
Add (1) and (2) =>2x+y+z = 137+241 = 378 --- 4
Subtract (1) and (2) => z-y = 241-137 = 104 ---- 5
Add (3) and (5) => y+z+2w+z-y = 200+104
=> 2(z+w) = 304
=> z+w = 152, A
Thanks
Anshu
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Anshu
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x^2-y^2=200^2 - 137^2;
x+y=241
(x-y)(x+y)=(200-137)(200+137)
x+y=241
x-y=200-137
x+y=241, 2x=304, x=152 --> CD=152
answer A
sanju09 wrote:The triangle ABC has sides AB = 137, AC = 241 an BC = 200. There is a point D, on BC, such that both incircles of triangles ABD and ACD touch AD at the same point E. What is the length of CD?
A. 152
B. 174
C. 179
D. 183
E. 197
[spoiler]https://gmatmaths.com/[/spoiler]
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