Question

Can anyone solve question #15

Answer 1

hope it will help you
thanks

Answer 2

Answer:

Step-by-step explanation:

Given: ∠ACB=90° and CD⊥AB.

To prove: $\frac{(BC)^2}{(AC)^2}=\frac{BD}{AD}$

Solution:

In ΔACD and ΔABC, we have

∠ADC=∠ACB=90°

∠CAD=∠BAC(Common)

Thus, by AA similarity,  ΔACD is similar to ΔABC.

Hence, $\frac{AC}{AB}=\frac{AD}{AC}$

⇒$(AC)^2=AB{\times}AD$                                (1)

Similarly, ΔDCB is similar to ΔCAB.

⇒$\frac{BC}{AB}=\frac{BD}{BC}$

⇒$(BC)^2=AB{\times}BD$                                 (2)

Now, dividing (2) by (1), we get

$\frac{(BC)^2}{(AC)^2}=\frac{AB{\times}BD}{AB{\times}AD}$

⇒$\frac{(BC)^2}{(AC)^2}=\frac{BD}{AD}$

Hence proved.