Mathematics
ABC is a triangle in which AB = AC and D is any point on BC. Prove that :
AB2 - AD2 = BD.CD
Pythagoras Theorem
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Answer
Given: ABC is a triangle in which AB = AC and D is any point on BC.
To prove: AB2 - AD2 = BD.CD
Construction: Draw AE ⊥ BC.
Proof: In Δ ABE and Δ ACE, we have:
AB = AC (Given)
AE = AE (Common)
∠AEB = ∠AEC (both are 90°)
Using RHS congruency criterion,
Δ ABE ≅ Δ ACE
⇒ BE = CE (by C.P.C.T.)
In Δ ABE, using Pythagorean theorem,
⇒ AB2 = AE2 + BE2 ……………….(1)
In Δ ADE, using Pythagorean theorem,
⇒ AD2 = AE2 + DE2 ……………….(2)
Subtracting equation (ii) from (i), we get:
⇒ AB2 - AD2 = (AE2 + BE2) - (AE2 + DE2)
⇒ AB2 - AD2 = AE2 + BE2 - AE2 - DE2
⇒ AB2 - AD2 = BE2 - DE2
⇒ AB2 - AD2 = (BE - DE)(BE + DE)
⇒ AB2 - AD2 = (BE - DE)(CE + DE) [∴ BE = CE]
⇒ AB2 - AD2 = BD.CD
Hence, AB2 - AD2 = BD.CD.
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