Question

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°. Prove that :
2AC² - AB² = BC² + CD² + DA²​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

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In quadrilateral ABCD, ∠B = 90° and ∠D = 90°. So, ΔABC and ΔADC are right-angled triangles.

In ∆ABC, using Pythagoras theorem,

AC2 = AB² + BC²

→AB² = AC² - BC²

In ∆ADC, using Pythagoras theorem,

AC² = AD² + DC²

LHS = 2AC² - AB²

= 2AC² - ( AC ² - BC² )

= 2AC² - AC² + BC²

= AC²+ BC²

= AD² + DC ² + BC²

= RHS