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Pythagoras Theorem

In Δ ABC, AB ...

Question

In $Δ A B C$ , AB = AC. Side BC is produced to D. Prove that $(A D^{2} - A C^{2}) = B D . C D$ .

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Solution

Const: Draw a perpendicular AE from A

Thus, AE ⊥BC

Proof:

In ΔABC, AB = AC

And AE is a bisector of BC

Then, BE = EC

In right angle triangles ΔAED and ΔACE

$A D^{2} = A E^{2} + D E^{2}$ —— (1)

$A C^{2} = A E^{2} + C E^{2}$ ——- (2)

Subtracting (2) from (1),

⇒ $(A D^{2} - A C^{2}) = D E^{2} - C E^{2}$

= $A D^{2} - A C^{2} = (D E + C E) (D E - C E) = (D E + B E) (D E - C E)$ [since CE = BE]

⇒ $A D^{2} - A C^{2} = B D \times C D$

Hence proved

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