Cyclic points

Let $\displaystyle A\equiv \left( 0,2 \right) \quad B\equiv \left( 2,3 \right) \quad C\equiv \left( 4,5 \right) \quad D\equiv \left( 0,t \right) .$

Slope of the line $\displaystyle AB={ m }_{ 1 }=\cfrac { 3-2 }{ 2-0 } =\cfrac { 1 }{ 2 }$ . Slope of the line $\displaystyle BC={ m }_{ 2 }=\cfrac { 5-3 }{ 4-2 } =1$ . If $\displaystyle \beta$ be the angle between $\displaystyle AB$ and $\displaystyle BC$ , then $\displaystyle \tan { \beta } =\cfrac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 2 }{ m }_{ 1 } } =\cfrac { 1-\cfrac { 1 }{ 2 } }{ 1+1.\cfrac { 1 }{ 2 } } =\cfrac { 1 }{ 3 }$ . Slope of the line $\displaystyle CD={ m }_{ 3 }=\cfrac { 5-t }{ 4-0 }$ .

Notice that $\displaystyle D\equiv \left( 0,t \right)$ lies on Y-Axis. So, if $\displaystyle \gamma$ be the angle of inclination of the line $\displaystyle CD$ with the X-Axis then clearly $\displaystyle { \tan { \gamma } =m }_{ 3 }$ . If $\displaystyle \alpha$ be the angle between $\displaystyle CD$ and $\displaystyle AD$ , then: $\alpha +\gamma =90°\\ \Rightarrow \cfrac { 1 }{ \tan { \left( \alpha +\gamma \right) } } =0\\ \Rightarrow \tan { \alpha } =\cfrac { 1 }{ \tan { \gamma } } =\cfrac { 4 }{ 5-t }$ .

Given that the four points are concyclic, means $\alpha +\beta =180\\ \Rightarrow \tan { \beta } =-\tan { \alpha } \\ \Rightarrow \cfrac { 1 }{ 3 } =\cfrac { 4 }{ t-5 } \\ \Rightarrow t=17$ . Hence the answer.