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Sum

In ∆PQR, PD ⊥ QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b)(a – b) = (c + d)(c – d).

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#### Solution

Given In ∆PQR, PD 1 QR, PQ = a, PR = b,QD = c and DR =d

To prove (a + b) (a-b) = (c + d)(c-d)

Proof In right angled ΔPDQ,

`PQ^2 = PD^2 + QD^2` .....[By Pythagoras theorem]

⇒ `a^2 = PD^2 + c^2`

⇒ `PD^2 = a^2 - c^2` .....(i)

In right-angled ∆PDR,

`PR^2 = PD^2 + DR^2` .......[By Pythagoras theorem]

⇒ `b^2 = PD^2 + d^2`

⇒ `PD^2 = b^2 - d^2` .....(ii)

From equations (i) and (iii),

`a^2 - c^2 = b^2 - d^2`

⇒ `a^2 - b^2 = c^2 - d^2`

⇒ `(a - b)(a + b) = (c - d)(c + d)`

Hence proved

Concept: Basic Proportionality Theorem (Thales Theorem)

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