What is the ratio of the area of triangle ABX to the area of triangle CDX?

Steps to Solve

1. Identify angles that are equal due to circle properties

Since A, B, C, and D lie on a circle, we can use the inscribed angle theorem. Angle $BAD = x$ is subtended by arc $BD$. Therefore, angle $BCD$ is also equal to $x$.

2. Identify vertically opposite angles

Angles $BCX$ and $BCD$ form a straight line, so they are supplementary angles, meaning they add up to $180^\circ$. Angle $DCX = y$. Therefore, $\angle BCD = 180^{\circ} - y$. Since $\angle BCD = x$, we have $x = 180^{\circ} - y$.

3. Express $y$ in terms of $x$

From the equation $x = 180^{\circ} - y$, we can express $y$ in terms of $x$ as $y = 180^{\circ} - x$.

4. Determine the angles of triangle CDX

We are given that angle $DCX = y$. We also know that $\angle CDX = \angle BDA$. Since angles $BDA$ and $BCA$ subtend the same arc $BA$, we have $\angle BDA = \angle BCA$. Angle $BAC$ and $BDC$ subtend the same arc $BC$ so $\angle BAC = \angle BDC$. We are given $\angle BAD = x$. We also know that the angles of quadrilateral $ABCD$ add up to $360^\circ$. Also, opposite angles in a cyclic quadrilateral sum to $180^\circ$. Therefore, $\angle BAD + \angle BCD = 180^\circ$ and $\angle ABC + \angle ADC = 180^\circ$. Since $\angle CDA + \angle ADX = 180^\circ$, we have $\angle ADX = 180^\circ - \angle CDA$. Since $\angle ABC + \angle ADC = 180^\circ$, and $\angle ABC + \angle ABX = 180^\circ$, we conclude that $\angle ADC = \angle ABX$. Therefore, $\angle ADX = \angle ABX$. Since $\angle ADX$ is also named $\angle CDX$ and are equal, we know that $\angle CDX = x$. So, in triangle $CDX$, we have $\angle DCX = y$, $\angle CDX = x$, and therefore $\angle CXD = 180^\circ - x - y$. Substituting $y = 180^\circ - x$, we have $\angle CXD = 180^\circ - x - (180^\circ - x) = 0^\circ$. This result is impossible for a triangle. Let's reconsider our approach

5. Correctly determine Angle CXD In quadrilateral $ABCD$, $\angle BCD + \angle BAD = 180^{\circ}$. Therefore $\angle BCD = 180 - x$. Since $BCX$ is a straight line, $\angle BCX + \angle BCD = 180^{\circ}$. Therefore, $\angle BCX = 180 - (180 - x) = x$. We are given $\angle DCX = y$. Therefore, $\angle BCX = x$. Now in triangle $BCX$, we have $\angle XBC + BCX + BXC = 180^{\circ}$. So $\angle XBC + x + \angle BXC = 180$. Therefore $\angle BXC = 180 - x - \angle XBC$.

Similarly, consider $\triangle ADX$. $\angle DAX = x$. So $\angle ADX + x + \angle DXA = 180^{\circ}$. Therefore $\angle ADX = 180 - x - \angle DXA$. Note that $\angle DXA = \angle BXC$. Let us call this angle $\alpha$. So, $\angle ADX = \angle ABC = 180 - x - \alpha$. Now in $\triangle ABX$, we have $\angle BAX + \angle ABX + \angle AXB = 180$. $\angle BAX = x$. So $x + 180 - x - \alpha + \alpha = 180$. This is always true.

Let's look at angles $CBX$ and $DAX$. Therefore $\angle CBX = \angle DAX = x$.

So we know angles $CBX= x$ and $DCX = y$. $\angle CXB = \angle DXA$. We also found that $\angle BCX = x$, so in $\triangle BCX, \angle BXC = 180 - 2x$. Therefore $\angle DXA = 180 - 2x$. Now in $\triangle ADX$, we have $\angle ADX = 180 - x - (180 - 2x) = x$. So $\angle ADX = x$. Similarly, $\angle CDX$.

So $\angle CDA = \angle CDX + \angle XDA$, and we have $\angle AXB = \angle CXD = 180 -2x$, $\angle BAD =\angle BCD =x$, $\angle ADC + \angle ABC = 180$. Now areas of similar triangles are proportional to the square of their corresponding sides.

6. Relate triangles ABX and CDX

Triangles ABX and CDX share a common angle at X. Area of ABX = $\frac{1}{2} * AX * BX * sin(AXB)$ Area of CDX = $\frac{1}{2} * CX * DX * sin(CXD)$ Since $\angle AXB = \angle CXD$, $\frac{Area(ABX)}{Area(CDX)} = \frac{AX * BX}{CX * DX}$

7. Use similarity

$\triangle ADX \sim \triangle BCX$ because $\angle DAX = \angle BCX$ and $\angle AXB = \angle CXD$ Thus, $\frac{AX}{CX}=\frac{DX}{BX}$ $AX * BX = CX * DX$ Therefore $\frac{AX * BX}{CX * DX} = 1$