In this diagram the center of the circle is A, mABD=20° and mDCA=52°. What is wrong with this diagram? Circle *A*

I know that:

mABD=20° -given

mDCA=52° -given

AB∥DC -given (from image)

ABACAD -all radiuses

AB and CD are cut by transversals AC and AB.

mBAO=52° -alternate interior angles

mADO=20° -alternate interior angles

I concluded that the thing that is wrong with this is diagram is that arcBC should be 104°, because of the inscribed angle theorem.

How do I know it isn't 104°? Am I even on the right track? Any explanation will be helpful.

  • 2
    $\begingroup$ $x = \angle{BDC} = \dfrac{1}{2}\angle{BAC}$, (angle made by chord $\overline{BC}$ on the arc is half the angle made at the centre) compare the angles of the triangles $\Delta BOA$ and $\Delta DOC$, $x+20 = 56+2x$ (absurd!) $\endgroup$ – r9m Dec 2 '15 at 4:05

If $AB \!\parallel\! CD,\ \angle{ABD}=\dfrac{1}{2}\angle{ACD}.$

Clearly, if $AD \!\parallel\! BC,\ \Delta ABC$ and $\Delta ACD$ are equilateral, and $\angle{ABD}=30 \hspace{.1em}^{\circ}\hspace{-.1em}.$ Alternatively, if $\angle{ABD}=20 \hspace{.1em}^{\circ}\hspace{-.1em},\ \angle{ACD}=40 \hspace{.1em}^{\circ}\hspace{-.1em}.$

As it is, if $A=\{0,0\}, \ D=\{1,0\},$ ie the quadrilateral is enclosed by the unit circle, $B=\{-\cos \left(2 \pi /9\right),-\sin \left(2 \pi /9\right)\}, \ C=\{\sin \left(7 \pi /90\right),-\cos \left(7 \pi /90\right)\},$ and $AB \!\not\parallel\! CD.$

If you have access to Mathematica, have a play with this:

With[{ε = 1/40}, Manipulate[With[{θ1 = (NSolve[-Cos[θa]/(1 - Sin[θa]) 
== Cot[θ] && 0 <= θa <= 2 Pi, θa] // FullSimplify)[[1, 1, 2]]}, 
Graphics[{Circle[{0, 0}, 1], Line@# & /@ {{{0, 0}, {Sin[θ], Cos[θ]}}, 
{{Sin[θ], Cos[θ]}, {1, 0}}}, Red, Line@# &/@ {{{0, 0}, {Sin[θ1], Cos[θ1]}}, 
{{Sin[θ1], Cos[θ1]}, {1, 0}}}, Black, Dashed, Line@# & /@ 
{{{0, 0}, {1, 0}}, {{Sin[θ], Cos[θ]}, {Sin[θ1], Cos[θ1]}}}, 
Red, PointSize[ε], Point@# & /@ {{Sin[θ], Cos[θ]}, {Sin[θ1], Cos[θ1]}, 
{0, 0}, {1, 0}}}, PlotRange -> {{-1 - 2 ε, 1 + 2 ε}, {-1 - 2 ε, 1 + 2 ε}}
]], {{θ, 23 Pi/18}, 0 + ε/10, 2 Pi - ε/10}]]

to get some intuition into how the angles behave when $AB \!\parallel\! CD.$


You cannot assume that AB||CD just because it looks like it is.

From the inscribed angle theorem you know BAC = 2BDC

From the fact that ABC is isosceles you know ABC=ACB and specifically ABC=90-BDC.

DBC is therefore 70-BDC.

You can now work that BDC is 32. BAC is therefore 58 degrees and AB is not parallel to CD.

  • $\begingroup$ Oops it was my mistake. The sheet did say that AB∥CD. Forgot to mention that. Thank you for the explanation. $\endgroup$ – Dana Dec 2 '15 at 4:23

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