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# In square ABCD, AD is 4 centimetres, and M is the midpoint of . Let O be the intersection of and . What is the ratio of OC to

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In square ABCD, AD is 4 centimetres, and M is the midpoint of $$\overline{CD}$$. Let O be the intersection of $$\overline{AC}$$ and $$\overline{BM}$$. What is the ratio of OC to OA? Express your answer as a common fraction.

Feb 25, 2020

#1
-1

By similar triangles, OC/OA = 1/4.

Feb 25, 2020
#2
+110227
+1

I think by similar triangles it is 1:2

Feb 25, 2020
#3
+111433
+2

Let A  = (0,4)    B  = ( 4,4)    C   =(4,0)   D   =  (0, 0)

The  line segment AC  lies on a line with an equation  of

y = -x + 4

AC  = 4sqrt (2)

The  segment MB   has a slope  of  [4 -0 ]  /[ 4 - 2]  =  4/2  = 2

So....this segment lies on a line with an equation of

y= 2 ( x  - 2)

y = 2x  - 4

The  x  coordinate of the  intersection of these  two lines is

-x + 4 =  2x  - 4

8 = 3x

x  = 8/3

And y  = -(8/3)  + 4 =   4/3

So  OC   =    sqrt  [  (4 - 8/3)^2  + (4/3)^2 ] = sqrt  [ ( 4/3)^2  + (4/3)^2  ] =  (4/3)sqrt (2)

And OA  =  AC  - OC  =  (4  - 4/3)sqrt (2) =   (8/3)sqrt (2)

So

OC             (4/3) sqrt (2)                  4               1

___  =       ____________  =         ___   =     _____

OA             (8/3) sqrt (2)                  8                2

Feb 25, 2020
edited by CPhill  Feb 25, 2020
#4
+111433
+3

Melody's  method  :

Triangle AOB  is  similar to triangle OCM

This means that

CM / AB  = OC / OA

CM  = (1/2)AB   .....

(1/2) AB          OC

_______  =     ___

AB              OA

1                 OC

__  =         ____

2                 OA

Feb 25, 2020
#5
+1133
+2

AD = DC = 4 cm

MC = DC/2 = 2

∠ACD = 45°

AC = AD / sin(45°) = 5.656854249

∠BMC = tan-1(BC/MC) = 63.435°

∠MOC = 180 - 45 - 63.435 = 71.565°

Let F be the foot of altitude from C to MO

CF = sin(∠BMC) * MC = 1.788854382

OC = CF / sin(∠MOC) = 1.885618645

OA = AC - OC = 3.771235604

The ratio of OC to OA = 1.885618645 / 3.771235604 = 0.5  or  1/2

Feb 25, 2020
edited by Dragan  Feb 25, 2020
#6
+110227
0

Yes I just did it in my head.

Using the rules of transversals cutting parallel lines it is easy to see that the triangles are similar.

MC is a half of AB

so

OC is a half of OA

1:2

Feb 26, 2020