Question

in the diagram, abcd is a cyclic quadrilateral. $overline{bc}$ and $overline{ad}$ are produced to e. $angle abc = 2r$, $angle cef = 5r$ and $angle dce = 51^circ$. find (i) the value of $r$; (ii) $angle adc$.

Explanation:

Step1: Recall cyclic quadrilateral property

In a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle. So, $\angle ABC = \angle DCE + \angle CDE$? Wait, no. Wait, when we produce $BC$ and $AD$ to $E$, $\angle ABC$ (interior angle of cyclic quadrilateral) and $\angle CDE$? Wait, actually, the exterior angle at $D$ (if we consider) but here, let's look at the angles. Wait, $\angle CEF$ is a straight - line related? Wait, $\angle CEF = 5r$ and the adjacent angle to it (linear pair) would be $180 - 5r$. But also, in cyclic quadrilateral $ABCD$, $\angle ABC$ (which is $2r$) and the angle $\angle ADC$ are related, but also, $\angle DCE = 51^{\circ}$ and the angle at $E$: Wait, let's use the property of cyclic quadrilaterals and exterior angles.

Wait, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. So, the angle $\angle DCE$ is an exterior angle to the cyclic quadrilateral $ABCD$ at vertex $C$. Wait, no, when we produce $BC$ to $E$ and $AD$ to $E$, so $\triangle CDE$: Wait, maybe we can use the fact that in triangle $CDE$, but also, the sum of angles on a straight line. Wait, $\angle CEF$ and the angle adjacent to it (let's call it $\angle CED$) are supplementary, so $\angle CED=180 - 5r$. Also, in triangle $CDE$, the sum of angles is $180^{\circ}$, so $\angle CDE+\angle DCE+\angle CED = 180^{\circ}$. But $\angle CDE$ is equal to $\angle ABC$ (because $ABCD$ is cyclic, and $\angle CDE$ is an exterior angle? Wait, no, $\angle ABC$ and $\angle ADC$ are supplementary (cyclic quadrilateral: opposite angles sum to $180^{\circ}$), and $\angle CDE$ is supplementary to $\angle ADC$ (linear pair), so $\angle CDE=\angle ABC = 2r$.

So in $\triangle CDE$: $\angle CDE+\angle DCE+\angle CED = 180^{\circ}$. We know $\angle CDE = 2r$, $\angle DCE = 51^{\circ}$, $\angle CED=180 - 5r$.

Substitute into the triangle angle - sum formula:

$2r+51+(180 - 5r)=180$

Step2: Simplify the equation

Simplify the left - hand side:

$2r+51 + 180-5r=180$

Combine like terms:

$(2r-5r)+(51 + 180)=180$

$-3r+231 = 180$

Subtract 231 from both sides:

$-3r=180 - 231$

$-3r=- 51$

Divide both sides by - 3:

$r=\frac{-51}{-3}=17$

Answer:

The value of $r$ is $\boldsymbol{17}$.