In figure given above, ABC is a right triangle right angled at C. If D is midpoint of BC, prove that AB² = 4AD² - 3AC²
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Given :
∠C = 90° and D is the mid-point of BC.
In ∆ ACD,
AD² = AC² + CD²
[By using Pythagoras theorem]
CD² = AD² - AC² ……….(1)
In ∆ACB,
AB² = AC² + BC²
[By using Pythagoras theorem]
AB² = AC² + (2CD)²
[D is the mid-point of BC]
AB² = AC² + 4CD²
AB² = AC² +4(AD² - AC²)
[from eq 1]
AB² = AC² + 4AD² - 4AC²
AB² = 4AD² - 4AC² + AC²
AB² = 4AD² - 3AC²
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