In quadrilateral ABCD, angle A = angle B / 2 = angle C / 3 = angle D / 4. Find all four angles.

Step-by-step solution

1. 1

   Turn the chain of equalities into one parameter. Because all four expressions are equal, give their common value a name: let $\widehat{A}=\frac{\widehat{B}}{2}=\frac{\widehat{C}}{3}=\frac{\widehat{D}}{4}=x.$ One unknown now controls all four angles.

2. 2

   Express every angle in terms of x. Multiplying each equation out: $\widehat{A}=x,\quad \widehat{B}=2x,\quad \widehat{C}=3x,\quad \widehat{D}=4x.$ So the angles are in the ratio $1:2:3:4$ - notice the denominators in the original statement become the multipliers.

3. 3

   Use the angle sum of a quadrilateral. A quadrilateral splits into two triangles, so its interior angles total $2\times180^\circ = 360^\circ$: $x+2x+3x+4x = 360^\circ.$

4. 4

   Solve for x. $10x = 360^\circ \quad\Longrightarrow\quad x = 36^\circ.$

5. 5

   Back-substitute. $\widehat{A}=36^\circ,\quad \widehat{B}=72^\circ,\quad \widehat{C}=108^\circ,\quad \widehat{D}=144^\circ.$

6. 6

   Check the total and the shape. $36+72+108+144 = 360$, as required. All four angles are less than $180^\circ$, so such a convex quadrilateral genuinely exists; note also that $\widehat{B}+\widehat{D} = 72+144 = 216^\circ \ne 180^\circ$, so it is not a cyclic quadrilateral.