Question

Here $△$ABC is a right angle triangle with right angle at C. Let BC =a , Ca = b, AB = c. Let P be the length of the perpendicular from C on AB. Prove that (i) Pc = ab.

Answer 1

In $△ACD,$

$A{D}^2+C{D}^2=A{C}^2$ (Using Pythagoras theorem)

$\Rightarrow A{D}^2=b^2-p^2$

$\Rightarrow AD=\sqrt{b^2-p^2}$--------------------(1)

In $△BCD,$

$B{D}^2+C{D}^2=B{C}^2$ (Using Pythagoras theorem)

$\Rightarrow B{D}^2=a^2-p^2$

$\Rightarrow BD=\sqrt{a^2-p^2}$-------------------(2)

In $△ABC$,

$A{C}^2+B{C}^2=A{B}^2$

$b^2+a^2=c^2$-------------------(3)

Now,

$AD+BD=AB=C$

$\Rightarrow \sqrt{b^2-p^2}+\sqrt{a^2-p^2}=c$

Squaring we get,

$\Rightarrow b^2-p^2+a^2-p^2+2\sqrt{b^2-p^2}\sqrt{a^2-p^2}=c^2$

$\Rightarrow a^2+b^2-2p^2+2\sqrt{b^2-p^2}\sqrt{a^2-p^2}=a^2+b^2$ $[\because c^2=a^2+b^2]$

$\Rightarrow p^2=\sqrt{b^2-p^2}\sqrt{a^2-p^2}$

Squaring,

$\Rightarrow p^4=a^2{b}^2+p^4-p^2(a^2+b^2)$

$\Rightarrow p^2(a^2+b^2)=a^2{b}^2$

$\Rightarrow p^2{c}^2=a^2{b}^2$ $(\because c^2=a^2+b^2)$

$\Rightarrow pc=ab$ (Taking the square root).