A trapezoid $ABCD$ $(AB\parallel CD)$ with base $AB=42$ circumscribes the circle $k(O;r)$. $k$ touches $AD$ at $Q$ and $AQ=18$ and $DQ=8$. What is the length of $CD$?
$AM = AQ= 18 ,BM = BN = 24 ,DQ = DP = 8 , NC = CP = x$.
$\triangle AOD$ and $\triangle BOC$ are right-angled.
$\triangle AOD \rightarrow QO^{2} = AQ\times QD \Leftrightarrow r^{2} = 18\times 8 \Rightarrow r = 12$. Can you explain to me why $QO^{2} = AQ\times QD$?
$\endgroup$2 Answers
$\begingroup$$QO^{2} = AQ\times QD $ is $\dfrac {QO}{AQ} = \dfrac {QD}{QO}$ in disguise. That set of ratio can be proved via similar trianlges. Within $\triangle AOD$ there are 3 trinagles that are similar to each other.
Another explanation is from a theorem called "power of a point".
$\endgroup$ $\begingroup$Because triangle $AOD$ is a right triangle with right angle at $O$ and $OQ$ is a perpendicular line from $O$ to the hypotenuse.
In fact, $AQ\times QD=CN\times NB= r^{2}$
You may want to read about right triangle
$\endgroup$
