Question

In a right angled $△ABC$ a circle is drawn with AB as a diameter which intersect hypotenuse $AC$ at point $P$. Prove that $PB=PC$

Answer 1

Let $\angle ABP$ be x & $\angle PBC=y$

then $x+y={90}^o$

$\angle PBC=\angle BAP$

(Alternate segment theorem)

$\angle APB={90}^o=x+y$

(Angle subtended by diameters ${90}^o$)

So $\angle BPC={90}^o$

In $\Delta PBC$

$\angle PBC+\angle BPC+\angle BCP={180}^o$ (A.S. p)

$y+90+\angle BCP={180}^o$

$\angle BCP=90-y$

$\angle BCP=x$

In $\Delta$PBC and ABC

$\angle P^2=\angle B$

$\angle C^2=\angle C$

$\angle B=\angle A$

so $\Delta PCB\sim \Delta BCA$

$\frac{PC}{BC}=\frac{CB}{AC}=\frac{PB}{AB}$

or $\frac{PC}{BC}=\frac{AB}{AC}=\frac{PB}{BC}$

so $\frac{PB}{PC}=\frac{BC}{BC}$

$\therefore \frac{PB}{PC}=1$

$\Rightarrow PB=PC$.