Points $E$ and $F$ are on side $BC$ of convex quadrilateral $ABCD$(with $E$ closer than $F$ to $B$). It is known that $\angle {BAE}=\angle {CDF}$ and $\angle {EAF}=\angle {FDE}$.Prove that $\angle {FAC}=\angle{EDB}$.

AEFD is a cyclic quadrilateral . But we need to show $ABCD is a cyclic quadrilateral but how? Please help me. Thank you!


As noted by you, we know that $AEFD$ is a cyclic quadrilateral, and so $\angle AEB = \angle ADF$.

We thus have that $$ \angle CBA + \angle ADC = \angle CBA + \angle ADF + \angle FDC = \angle CBA + \angle BAE + \angle AEB = 180^\circ. $$

Thus $ABCD$ is cyclic.



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.