Question

∆ABD is a right triangle in which ∠A = 90° and AC ⟂ BD.

$Prove \: that$
1.) AB² = BC.BD

2.) AC² = BC.DC

3.) AD² = BD.CD​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$\large\textsf{Given:}$

★ ∠A = 90°

★ AC ⊥ BD

(i)$\large\textsf{To Find:}$

⇒ AB² = BC. BD

$\large\textsf{Solution:}$

In ∆BCA and ∆BAD,

⇒ ∠BCA = ∠BAD ---- Each angle 90°

⇒∠B = common

So, ∆BCA ∽ ∆BAD ----AA Similarity -(1)

Hence, $\large\frac{BC}{AB} = \frac{AC}{AD} = \frac{AB}{BD}$ ----C.S.S.T

And, ∠BAC = ∠BDA ----C.A.S.T -(2)

So, $\large\frac{BC }{AB} = \frac{AB}{BD}$

∴ (AB)² = BC. BD

Hence Proved!

(ii) $\large\textsf{To Find:}$

⇒ AC² = BC. DC

$\large\textsf{Solution:}$

In ∆BCA and ∆DCA,

⇒ ∠BCA = ∠DCA ---Each angle 90°

⇒ ∠BAC = ∠CDA ---from (2)

So, ∆BCA ~ ∆ACD ----AA Similarity -(3)

Hence, $\large\frac{BC}{AC} = \frac{AC}{CD} = \frac{AB}{AD}$ ---C.S.S.T

So, $\large\frac{BC}{AC} = \frac{AC}{CD}$

∴ AC² = BC. DC

Hence proved!

(iii) $\large\textsf{To Find:}$

⇒ AD² = BD. CD

$\large\textsf{Solution:}$

From (1) and (2), we get-

⇒ ∆BAD ~ ∆ACD

Hence, $\large\frac{AB}{AC} = \frac{AD}{CD} = \frac{BD}{AD}$

So, AD² = BD. CD

Hence Proved!