Quadrilateral $ABCD$ is cyclic, with $BC=BD=1$. If $AD:AC:AB=1:7:5$, then find $AD^2$. I tried applying Ptolemy's theorem, I got $CD$, and then some angles but it did not give me something.

  • $\begingroup$ If you got CD, than the rest should be easy... I think. $\endgroup$ – Moti Mar 24 '15 at 5:37
  • $\begingroup$ How? What should I do then? $\endgroup$ – user167045 Mar 24 '15 at 5:50
  • $\begingroup$ Did you try Heron's formula for triangle area? $\endgroup$ – Moti Mar 24 '15 at 5:59
  • $\begingroup$ Did you try this? - If ABCD is a cyclic quadrilateral, then AC⋅(AB⋅BC+CD⋅DA)=BD⋅(DA⋅AB+BC⋅CD) $\endgroup$ – Moti Mar 24 '15 at 6:01
  • $\begingroup$ Thanks, but how did you get this result? Also, from this I am getting the answer 6/192, could you check, where is it wrong? $\endgroup$ – user167045 Mar 24 '15 at 6:12

Let $AD=x=\frac{AC}{7}=\frac{AB}{5}$. Then according to Ptolemy's theorem, $1(7x)=1(x)+(CD)(5x)$, which gives $CD=\frac{6}{5}$. Now let the midpoint of $CD$ be $M$. We see that $\cos\angle CAB=\cos\angle CDB=\frac{DM}{BD}=\frac{3}{5}$, and then it follows from the law of Cosines on triangle $ABC$ that: $$(5x)^2+(7x)^2-2(7x)(5x)(\frac{3}{5})=1^2\implies x^2=\boxed{\frac{1}{32}}$$Geo Diagram


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