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# Geometry Math Question HELP!!!

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ABCD is a parallelogram such that $$AB=\sqrt{3}\cdot BC.$$ Diagonals   $$\overline{AC}$$ and $$\overline{BD}$$  intersect at $$X$$ and $$\angle ABC=\angle BXC.$$Find the measure of $$\angle{ACB}$$ in degrees.

~~~Thanks~~~

Dec 8, 2019
edited by RandomPerson  Dec 9, 2019
edited by RandomPerson  Dec 9, 2019

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Angle ACB  =  Angle XCB

Angle ABC  = Angle BXC

So.....by AA congruency......triangle CBA  is similar to triangle  CXB

Therefore

AB/ BC  =√3/ 1  = BX / XC ⇒  BX = √3 *  XC   ⇒   (BX)^2  = 3(CX)^2

And  by a property of  parallelograms,   AC  = 2CX

And

CX/ BC  =  BC / CA

BC^2  =  CA * CX

BC^2  =  2XC * CX

BC^2  = 2(CX)^2

BC  = √2 * CX

So .....using the Law of Cosines

BC^2  =  BX^2  + CX^2  -  2(BX)(CX) cos (CXB)

2(CX)^2  = 3(CX)^2 + (CX)^2  - 2(√3*CX * CX)  cos (CXB)

-2(CX)^2  =    -  2√3 * (CX)^2 cos (CXB)

1 / √3  =  cos (CXB)

So  the sin  of  CXB   =    √ [ 3 - 1]/ √3   =    √(2/3)

So....by the Law of Sines

sin XCB  / BX  =  sin (CXB) / BC

sin XCB / (√3 CX)  =  √(2/3) / (√2 CX)

sin XCB =   √3 * (√2/ √3)

_________     =      1

√2

sin (XCB)  =  1  = sin ACB

arcsin (1)  =  ACB  =  90°   Dec 9, 2019
edited by CPhill  Dec 9, 2019