+12 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~~~
+129847 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°
