Question

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In ∆ ABC, seg AD is parallel to seg BC
Prove that AB²+CD²= BD²+AC²Q
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Answer 1

In △ADC, ∠ADC=90°

∴ AC² =AD² + CD² (By Pythagoras theorem) ....

(1)

In △DBA, ∠ADB= 90°

∴ AB² = AD² + BD² (By Pythagoras theorem) ....

(2)

Subtracting (1) from (2), we get

AB² – AC² = AD² + BD² – AD² – CD²

∴ AB² – AC² = BD² – CD²

∴ AB² + CD² = BD²+AC²

Answer 2

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In ∆ADC,∠ADC = 90°

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎∴AC² = AD²+ CD²(By Pythagoras theorem). . . . . . . .(1)

In ∆DBA,∠ADB = 90°

‎ ‎ ‎ ‎ ‎ ‎ ‎∴AB²= AD²+BD²(By Pythagoras theorem). . . . . . . .(2)

Subtracting (1) from (2), we get

AB² - AC² = AD² + BD² - AD² - CD²

‎ ‎ ‎ ‎ ‎ ‎ ‎∴AB² - AC² = BD² - CD²

‎ ‎ ‎ ‎ ‎ ‎ ‎∴AB² + CD² = BD² + AC²

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