In triangle ABC if AP perpendicular …

According to question it is given that in ∆ ABC,
AC² = BC² - AB²
{tex}\Rightarrow{/tex} AC² + AB² = BC²
{tex}\therefore \triangle{/tex}ABC is right angled at A (by using converse of Pythagoras theorem)
{tex}\therefore {/tex} {tex}\angle{/tex} A = 90°
Now, In {tex}\triangle{/tex}APC
{tex}\angle{/tex}PAC + {tex}\angle{/tex}C = 90° ........(i)
Similarly, {tex}\angle{/tex}B + {tex}\angle{/tex}C = 90° .........(ii)
from (i) and (ii), we get
{tex}\angle{/tex}B + {tex}\angle{/tex}C = {tex}\angle{/tex}PAC + {tex}\angle{/tex}C
 {tex}\Rightarrow{/tex} {tex}\angle{/tex}B = {tex}\angle{/tex}PAC
In {tex}\triangle{/tex}BPA and {tex}\triangle{/tex}APC,
{tex}\angle{/tex}B = {tex}\angle{/tex}PAC (proved)
{tex}\angle{/tex}BPA = {tex}\angle{/tex}APC (each 90°)
{tex}\therefore \quad \triangle B P A \sim \triangle A P C{/tex}
{tex}\Rightarrow \quad \frac { \mathrm { PB } } { \mathrm { PA } } = \frac { \mathrm { PA } } { \mathrm { CP } }{/tex}
{tex}\Rightarrow \quad \mathrm { PB } \times \mathrm { CP }= PA^2{/tex}
Hence proved.