touch AD at the same point E

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Wed Mar 09, 2011 3:24 am

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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Source: Beat The GMAT — Problem Solving | Wed Mar 09, 2011 3:24 am

anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Wed Mar 09, 2011 10:57 am

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]

Please refer to the diagram below :
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

(Every mistake is a lesson learned )

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Wed Mar 09, 2011 1:49 pm

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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