Question

Please answer the given questions :

1. In Triangle ABC AD is perpendicular to BC. (AD)2= BD X CD. prove that AB2+AC2=(BD +CD) 2.


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Answer 1

Answer:

In the given Fig., AD ⊥ BC. Prove that AB2 + CD2 = BD2 + AC2.

In right ∆ADC. by Pythagoras Theorem,

AC2 = AD2 + CD2 ...(i)

In right ∆ADB, by Pythagoras Theorem,

AB2 = AD2 + BD2 ...(ii)

Subtracting (i) from (ii), we get

AB2 - AC2 = BD2 - CD2

⇒ AB2 + CD2 = BD2 + AC2 Hence Proved.

Answer 2

$\huge{\mathfrak{Solution:-}}$

★ ${\mathtt{\underline{Given:}}}$ AD is drawn perpendicular to BC.

★ ${\mathtt{\underline{To\;Prove:}}}$ AB² - AC² = BD² - CD²

When AD is drawn perpendicular to BC, two triangles ADB and ADC are obtained.

Considering triangle ADB,

AB² = AD² + BD²     ................(1)   (using Pythagoras theorem)

Considering triangle ADC,

AC² = AD² + CD²    ................(2)  (using Pythagoras theorem)

Subtracting equation (2) from equation (1),

⇒ AB² - AC² = AD² + BD² - (AD² - CD²)

⇒ AB² - AC² = AD² + BD² - AD² - CD²

⇒ AB² - AC² = BD² - CD²

Hence Proved