1
$\begingroup$

Let $ABCD$ be a cyclic convex quadrilateral such that $AD + BC = AB$. Prove that the bisectors of the angles ADC and BCD meet on the line $AB$.enter image description here

I tried to find similar triangles since the angles are getting bisected, however I couldn't go anywhere in that direction. I also tried seeing if there were any properties that could be useful about the cyclic quadrilateral. I found properties from here: https://www.quora.com/What-are-the-properties-of-a-cyclic-quadrilateral-with-images

$\endgroup$
  • 1
    $\begingroup$ Please read descriptions of tags before using them - you used two tags (algebraic-geometry and euclidean-algorithm) which specifically said in their description that they were not applicable to this sort of problem. $\endgroup$ – KReiser Feb 5 at 0:10
3
$\begingroup$

enter image description here

Let angle bisector for $\angle BCD$ meet $AB$ at $F$ (so we have to prove that $DF$ is angle bisector for $\angle ADC$), then $$\angle BCF = FCD = \alpha\;\;\;\;{\rm and }\;\;\;\;\;\angle BAD = 180^{\circ} -2\alpha$$ and let $E$ on $AB$ be such that $BE = BC$, then $AE = AD$ and $$\angle FED = \angle AED = \angle ADE = \alpha$$ so $\angle FED = \angle FCD = \alpha $ and thus $CDFE$ is cyclic!

Now, if $\angle FDC = \beta$ then $\angle BEC = \angle ECB = \beta$, so $$\angle 180^{\circ} -2\beta \implies \angle ADC = 2\beta$$ and thus $DF$ is also angle bisector but for $\angle ADC$ and we are done.

$\endgroup$
  • 1
    $\begingroup$ is angle FDE equal to angle beta or is angle FDC equal to angle beta? is there a mistake on the diagram? $\endgroup$ – user604720 Feb 10 at 1:56
  • $\begingroup$ Second one, i made an edit. $\endgroup$ – Aqua Feb 10 at 11:03
1
$\begingroup$

Let $\angle D=2\delta$ and $\angle C=2\gamma$, and wlog assume $\delta\geqslant \gamma$ (to be consistent with your picture). Then $\angle A=180^\circ-2\gamma$ and $\angle B=180^\circ-2\delta$. Denote by $M$ the point on $AB$ such that $AM=AD$, so $BM=BC$ as well. Then $\angle ADM=\angle AMD=\gamma$ and $\angle BMC=\angle BCM=\delta$.

Denote by $N$ the intersection of $AB$ and the bisector of $\angle D$; since $\delta\geqslant\gamma$ we have $\mathcal B(A,M,N,B)$, and $\angle MDN=\delta-\gamma$. Since $\angle NMC=\angle BMC=\delta$ and $\angle NDC=\delta$ we get that $NCDM$ is cyclic, so $\angle MCN=\angle MDN=\delta-\gamma$. Now, $\angle BCN= \angle BCM-\angle NCM= \delta-(\delta-\gamma)= \gamma$, which means that $N$ is on the bisector of $\angle C$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.