Question

in the right angled triangle ABC angleABC=90° segBD hypotenuse AC rhen BD² is equal to​

Answer 1

Answer:

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BC = 90°

BD perpendicular AC

in A ADB & AABC

ADB = ABC = 90°

A = A (common)

=> A ADB = AABC

=> AD/AB = BD/BC=AB/AC

AD/AB = BD/BC

=> BD = AD * BC/AB Eq1

in A BDC & AABC

BDC ABC = 90°

C = C (common)

=>ABDC = AABC => BD/AB=BC/AC = CD/BC

BD/AB = CD/BC

=> BD = AB CD /BC Eq 2

Eq1* Eq2

=> BD² = (AD * BC/AB) * (AB* CD/BC)

=> BD² = AD * CD

Answer 2

Step-by-step explanation:

Given:

∠ABC=90o,BD⊥AC,

BD=8 and AD=4 as in diagram.

Let ∠ACB=x

⇒∠CBD=90−x

⇒∠ABD=x

⇒∠BAD=90−x

⇒ΔABC,ΔADB and ΔBDC are similar

⇒ BD/CD= AD/DB

⇒ 8/CD=4/8

⇒CD= 8×8/4 =16cm.

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