In triangle ABC , seg AD ⊥ seg BC, DB = 3CD. Prove that: 2AB² = 2AC² + BC² Figure: D lies on side BC with C–D–B.

Answer: With BC = 4CD, Pythagoras theorem gives AB² - AC² = 8CD² = BC²/2 , hence 2AB² = 2AC² + BC² . BC = CD + DB = CD + 3CD = 4CD, so BC² = 16CD² . In right-angled triangle ADB : AB² = AD² + DB² = AD² + 9CD² In right-angled triangle ADC : AC² = AD² + CD² Subtracting, AB² - AC² = 8CD² 2AB² - 2AC² = 16CD² = BC² therefore 2AB² = 2AC² + BC² Hence proved.

In ΔABC, seg AD ⊥ seg BC, DB = 3CD. Prove that: 2AB² = 2AC² + BC²

In \(\triangle ABC\), seg AD ⊥ seg BC, DB = 3CD. Prove that: \(2AB^2 = 2AC^2 + BC^2\)

Figure: D lies on side BC with C–D–B.

Doubtly Asked 24 Oct 2026 Save WhatsApp Report

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Answer: With BC = 4CD, Pythagoras theorem gives \(AB^2 - AC^2 = 8CD^2 = \frac{BC^2}{2}\), hence \(2AB^2 = 2AC^2 + BC^2\).

BC = CD + DB = CD + 3CD = 4CD, so \(BC^2 = 16CD^2\).

In right-angled \(\triangle ADB\):

\[AB^2 = AD^2 + DB^2 = AD^2 + 9CD^2\]

In right-angled \(\triangle ADC\):

\[AC^2 = AD^2 + CD^2\]

Subtracting,

\[AB^2 - AC^2 = 8CD^2\] \[2AB^2 - 2AC^2 = 16CD^2 = BC^2\] \[\therefore 2AB^2 = 2AC^2 + BC^2\]

Hence proved.

Doubtly AI · 24 Oct 2026 AI-verified · Oct 2026 Report

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In ΔABC, seg AD ⊥ seg BC, DB = 3CD. Prove that: 2AB² = 2AC² + BC²

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