In triangle abc,AB=AC, BD perpendicular to AC. Show that db square minus cd square is 2× cd × ad

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Solution

Given: AB=AC , BD _|_ AC

To Prove: BD² -DC² = 2DC × AD

Proof: BC² =CD^{2 +} BD² [PYTHA. THEO.]

=> BD² =BC²-CD²

AB² = AD²+ BD² [PYTHA. THEO.]
=> BD² =AB²-AD²

=[ AB+AD ] × [ AB-AD ] {A²-B² = [A+B] [A-B] }

=[ AC+AD ] [ AC-AD] { Since AB=AC}

=[ AC+AD ] × DC

=AC×DC + AD×DC

=[AD+DC]×DC + AD×DC
{Since AC = AD+DC}

=AD×DC+DC²+AD×DC

BD² = 2AD*DC+DC²

BD² - DC² = 2DC×AD

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