Question

Can anyone solve this question

Answer 1

Now, in ∆ADP and ∆CBP,
∠ADP=∠CBP (angle in the same segment)
∠APD=∠CPB (vertically opposite angles)
therefore, ∆ADP ~ ∆CBP (AA Similarity)
so,
$\frac{DP}{BP} = \frac{AP}{PC} \\ = > \frac{1}{x} = \frac{x}{8} \\ = > {x}^{2} = 8 \\ \: or \\ x = 2 \sqrt{2}$

Please mark my answer as brainliest.