In \(\triangle PQR\), let \(PD\perp QR\) with \(D\in QR\). If \(PQ=a\), \(PR=b\), \(QD=c\) and \(DR=d\), prove
\[(a+b)(a-b)=(c+d)(c-d).\]
Proved.
Step 1: Look at right triangle \(\triangle PQD\).
Here, \(PQ = a\) is the hypotenuse, \(PD\) is one leg, and \(QD = c\) is the other leg.
\(PQ^2 = PD^2 + QD^2\)
So, \(a^2 = PD^2 + c^2.\)
Step 2: Look at right triangle \(\triangle PRD\).
Here, \(PR = b\) is the hypotenuse, \(PD\) is one leg, and \(DR = d\) is the other leg.
By Pythagoras theorem:
\(PR^2 = PD^2 + DR^2\)
So, \(b^2 = PD^2 + d^2.\)
Step 3: Subtract the two equations.
From Step 1: \(a^2 = PD^2 + c^2\)
From Step 2: \(b^2 = PD^2 + d^2\)
Subtracting,
\(a^2 - b^2 = (PD^2 + c^2) - (PD^2 + d^2)\)
\(a^2 - b^2 = c^2 - d^2.\)
Step 4: Use difference of squares formula.
\(a^2 - b^2 = (a+b)(a-b)\)
\(c^2 - d^2 = (c+d)(c-d)\)
So, \((a+b)(a-b) = (c+d)(c-d).\)
Therefore proved.