In triangle ABC, angle C = 90° and CD is perpendicular bisector of AB then
BC square/AC square ???
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Answer:
In
Δ CAD, CA² = CD² + AD² .... (1)
Also in
Δ CDB, CB² = CD² + BD² .... (2)
(eqⁿ 1) + (eqⁿ 2) we get,
CA² + CB² = 2CD² + AD² + BD²
→ AB² = 2CD² + AD² + BD²
→ AB² - AD² = BD² + 2CD²
→ (AB + AD)(AB - AD) - BD² = 2CD²
→ (AB + AD)BD - BD² = 2CD²
→ BD(AB + AD - BD) = 2CD²
→ BD(AD + AD) = 2CD²
→ BD × 2AD = 2CD²
→ CD² = BD × AD
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