Question

In a quadrilateral ABCD, ∟B = ∟D = 90◦ Prove that:

2AC² - BC² = AB² + AD² + DC²

Answer 1

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Answer 2

Given :-

• A quadrilateral ABCD.
• ∟B = ∟D = 90◦

To prove :-

• 2AC² - BC² = AB² + AD² + DC²

Solution :-

• In ∆ADC

→ AC^2 = AD^2 + CD^2

→ 2 × AC^2 = 2(AD^2 + CD^2) eqaution(1) [ by Pythagoras theorem ]

• In ∆ABC

→ BC^2 = AC^2 - AB^2 equation(2)

• Equation (1) - equation(2)

→ 2AC^2 - BC^2 = 2( AD^2 + CD^2) - AC^2 + AB^2.

→ 2AC^2 - BC^2 = 2AD^2 + 2CD^2 - AC^2 + AB^2.

→ 2AC^2 - BC^2 = 2AD^2 - AC^2 + CD^2 + CD^2 + AB^2.

→ 2AC^2 - BC^2 = 2AD^2 - (AC^2 - CD^2) + CD^2 + AB^2.

→ 2AC^2 - BC^2 = 2AD^2 - AD^2 + CD^2 + AB^2.

→ 2AC^2 - BC^2 = AD^2 + CD^2 + AB^2.

Hence proved !