Question

solve the question please.... ​

Answer 1

In ∆KPQ and ∆KLM;

angle KQP = angle KML ... (each 45°, given)

angle PKQ = angle LKM ... (common angle)

∆KPQ ~ ∆KLM ... (by AA test of similarity)

$\frac{KQ}{KM} = \frac{PQ}{LM}$

(corresponding sides of similar triangles are proportional)

$\frac{c}{b + c} = \frac{x}{a} \\ \\ x = \frac{ac}{b + c}$