Question

In the given fig. if AD perpendicular to BC . Prove that
${ab }^{2} + \: cd {}^{2} \: = bd { }^{2} \: + ac {}^{2}$


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Answer 1

Step-by-step explanation:

From Δ ADC, we get :

⇒ AC² = AD² + CD² - i)

∵ Pythagoras theorem

From Δ ADB, we get :

⇒ AB² = AD² + BD² - ii)

∵ Pythagoras theorem

Subtract i) from ii) :

⇒ AB² - AC² = BD² - CD²

⇒ ∴ AB² + CD² = BD² + AC²

Answer 2

Step-by-step explanation:

SOLUTION ✍️

_____________________

From Δ ADC, we get :

⇒ AC² = AD² + CD² - i)

∵ Pythagoras theorem

From Δ ADB, we get :

⇒ AB² = AD² + BD² - ii)

∵ Pythagoras theorem

Subtract i) from ii) :

⇒ AB² - AC² = BD² - CD²

⇒ ∴ AB² + CD² = BD² + AC²

hope this helps you!!