Question

hey...

50 points with clear explanation plz...

Answer 1

HEY MATE HERE YOUR ANSWER--

∆ACD~∆ABC ( BY theorem 6.7)

$\frac{ac}{ab } = \frac{ad}{ac}$

AC²=AB . AD

∆BCD~ BAC ( theorem 6.7)

$\frac{bc}{ba} = \frac{bd}{bc}$
BC²=BA . BD

BY THIS

$\frac{ {bc}^{2} }{ {ac}^{2} } = \frac{ba.bd}{ab.ad} = \frac{bd}{ad}$

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MATE'S

Answer 2

Answer:

BC²/AC² = BD/AD

Step-by-step explanation:

From figure:

(i)

ΔACD ~ ΔABC

⇒ AC/AB = AD/AC

⇒ AC² = AB * AD  

(ii)

ΔBCD ~ ΔBAC

⇒ BC/BA = BD/BC

⇒ BC² = BA * BD

From (i) & (ii), we get

BC²/AC² = (BA * BD)/(AB * AD)

              = BD/AD.

∴ LHS = RHS

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