Amir Hossein wrote: In the quadrilateral $ABCD$, we have $\measuredangle BAD = 100^{\circ}$, $\measuredangle BCD = 130^{\circ}$, and $AD=1$ centimeter. Find the length of diagonal $AC$. Not enough data given!!

sunken rock wrote: Not enough data given!! How about now? Thanks, fixed.

Amir Hossein wrote: In the quadrilateral $ABCD$, we have $\measuredangle BAD = 100^{\circ}$, $\measuredangle BCD = 130^{\circ}$, and $AB=AD=1$ centimeter. Find the length of diagonal $AC$. I think it is still not enough. If we fix $A,B,D$, the point $C$ can be any point on some circle.

MeineMeinung wrote: Amir Hossein wrote: In the quadrilateral $ABCD$, we have $\measuredangle BAD = 100^{\circ}$, $\measuredangle\color{red} {BCD} = 130^{\circ}$, and $AB=AD=1$ centimeter. Find the length of diagonal $AC$. I think it is still not enough. If we fix $A,B,D$, the point $C$ can be any point on some circle.

Answer is 1 cm

yes I think I is 1cm

If you Stretch the altitude of ABC triangle to cut BC and call E new point ADCE is Cyclic quadrilateral and AEC=ACD , AED=ADC and we know AEC=AED so AD=AC=1

let $O$ the circumcircle of $BCD $ , we have $\angle BOC=100^\circ $ hence $BCAO$ are cyclic but $O,A$ li e on the $BC$-bisector then $A=O$ implies $AC=1$

Draw the circle $C(A,AB)$; any point $X$ onto the bigger arc $\stackrel{\frown}{BD}$ has $m(\widehat{BXD})=50^\circ$, that is, $XBCD$ is cyclic, wherefrom $AC=AB$. Best regards, sunken rock

We have $$ \angle ABC + \angle CDA = 130^{\circ} = \angle BCD$$so $$ (\angle ABC - \angle BCA ) + (\angle CDA - \angle ACD) = 0$$But the LHS has the same sign as that of $$ (AC - AB) + (AC - AD) = 2(AC-1)$$so that $AC = 1$.

We can also denote some $A_{1}$ such that $A_{1}BA$ is isosceles with angle $BA_{1}D$=50° Then notice that $A_{1}BCD$ is cyclic and $A_{1}BD$=90° and then get that $A$ is the center of the circumscribed circle. From here it is obvious that $AC$ is the radius $AC=AD=1$

This is from the competition of the Tournament Of Towns -1984-85

WLOG let $BC>CD$ (the case $BC<CD$ is analogous). Let the perpendicular bisector of $BD$ intersect $BC$ at $L$ (this is due to $BC>CD$ in $\triangle BCD$). Since $\triangle BAD$ is isosceles, the perpendicular bisector will pass through $A$. We now have $DL=BL$ and $AD=AB$, so $\triangle ABL \cong \triangle ADL$. Also, $\angle BAL=50$ and if we let $\angle ABC=\phi$, we will get $\angle BLA=130-\phi$, but $\angle ADC=130-\phi$ too, so $ALCD$ is cyclic. Now we have $\angle ABC=\angle ADL=\angle ACL$, so $\triangle ABC$ is isosceles and $AC=AB=1$. Now in the case $BC=CD$ we directly get that $ABCD$ is a deltoid, so $\angle ADC=65$, while $\angle ACD=130/2=65$ and so $AC=AD=1$.