Chapter 9.5, Problem 27WE

To determine

ToProve :In an inscribed quadrilateral $AB·CD+BC·AD=AC·BD$ .

Explanation of Solution

Giveninformation:

  

Inscribed angle $\angle CDB=\angle CAB$

Consider the following figure:

  

Inscribed angle, $\angle CDB=\angle CAB$ ,

On $DC,\angle DAC=\angle DBC$

Take Q such that, $\angle DCQ=\angle BCA$

Since, $NADCQ+\angle BCQ=\angle DCB=\angle DCA+\angle BCA$

Therefore, $\angle BCQ=\angle DCA$

  $△DCQ\simeq △ACB$

Similarly, $△DCA\simeq △QCB$

  $\begin{array}{l}\frac{DQ}{DC}=\frac{BA}{CA}\hfill \\ \frac{BQ}{CB}=\frac{AD}{CA}\hfill \end{array}$

Add the two equations will give,

  $DB·CA=AB·CD+AD·CB$