Question

In the adjoining figure, $BC$ is a diameter of the circle with centre $M.PA$ is a tangent at $A$ from $P$, which is a point on line $BC$. $AD\perp BC$. Prove that $D{P}^2=BP\times CP-BD\times CD$

Answer 1

Given: In the given figure $BC$ is the diameter of the circle with centre $M,PA$ is the tangent at $A$ from $P$ which is a point on line $BC$ and $AD\perp BC$
To Prove:
$D{P}^2=BP\times CP-BD\times CD$
Construction: Join seg $AB$ and seg $AC$
Proof: In $△ABC,\angle BAC={90}^{\circ }$ (angle subtended by diameter in semi circle)
seg $AD\perp$ side $BC$ ....(Given)
$\therefore$ By property of geometric mean,
$A{D}^2=BD\times CD$ ... (i)
Ray $PA$ is a tangent and $PB$ is a secant
$\therefore$ By tangent secant theorem,
$P{A}^2=BP\times CP$ ... (ii)
In right angled $△PAD$, the Pythagoras theorem,
$P{A}^2=D{P}^2+A{D}^2$
$\therefore D{P}^2=P{A}^2-A{D}^2$ .... (iii)
From (i), (ii) and (iii), we get
$D{P}^2=BP\times CP-BD\times CD$