There was a question in geometry which was marked as a simple question as follows(Diagram below):-

The diagonal AC of a quadrilateral $CDAB$ bisects the $\angle DCB$. It is given that $BC=BD$. $\angle CAB = 80$, $\angle DBC = 20$, then the measure of $\angle ADC$ in degree is......

Lets take the point of intersection of the two diagonals as O. Now since $\triangle DCB$ is isosceles, $\angle D$ (Of the triangle) + $\angle C + 20 = 180$. Simplifying, we get $\angle CDA = 80$, and thus $\angle ACD$ and $\angle ACB = 40.$ Now solving for $ \triangle ACB$ (Taking the whole of $\angle C$ as $20+x$) we get $x = 40$. Now $\angle O$ in $\triangle AOC = 120$ and using vertically opposite angles formula, i got the value of $\angle AOB$ and $DOC = 60$. Now i'm stuck here. From my diagram i can see that $\angle CAB$ and $\angle BDC$ are equal to each other And $\angle ABD$ and $\angle DCA$ are equal to each other. I thought of assuming $\angle BDA = \angle ACB$ but i have no formal proof for that.

ANY HELP IS APPRECIATED enter image description here


As $\angle CDB=\angle CAB$, quadrilateral $ABCD$ is cyclic. It follows that $\angle ADB=\angle ACB=40°$.

  • $\begingroup$ Thanks for the response. But can you please give me a proof why the quadilateral is cyclic? $\endgroup$ – infixint943 Dec 10 '17 at 17:46
  • $\begingroup$ Because if two equal angles insist on the same chord, then they belong to the same circular arc. $\endgroup$ – Aretino Dec 10 '17 at 17:49
  • $\begingroup$ I'm sorry, but I didn't get you. Can you please elaborate or at least send me a link to the explanation $\endgroup$ – infixint943 Dec 11 '17 at 9:45
  • 1
    $\begingroup$ See here for the inverse of the corollary of the inscribed angle theorem: cut-the-knot.org/pythagoras/Munching/inscribed.shtml $\endgroup$ – Aretino Dec 11 '17 at 12:57

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