Question

in triangle ABC angle C= 90 degrees and CD is perpendicular to AB. prove that BC2/AC2= BD/AD​

Answer 1

Answer:

Step-by-step explanation:

To prove CD² = BD × AD

In Δ CAD, CA² = CD² + AD² .... (1)

Also in Δ CDB, CB² = CD² + BD² .... (2)

(1) + (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

Hence proved.

Answer 2

Answer:

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Step-by-step explanation:

To prove CD² = BD × AD,

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 ,

Hence proved !!!!.

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