Question

plz help me with good explanation ( second part) of qstn no 25.

Answer 1

Given:-
In the triangle ABC, ∠A is obtuse
 and PB⊥ AC and QC⊥ AB

Consider triangles BPA and AQC
∠P = ∠Q = 90°
∠PAB = ∠QAC [Vertically opposite angles]
=> ΔBPA ~ ΔAQC

=> AB/AC = AP/AQ
=> AB x AQ = AP x AC  ..............1

Now, consider, triangle BCQ
  BC2 = CQ2 + BQ2                          [By Pythagoras theorem]
=>BC2 = CQ2 + (AB + AQ)2                    [ BQ = AB + AQ]
=>BC2 = [CQ2 + AQ2] + AB2 + 2AB × AQ2

In triangle ACQ,
CQ2 + AQ2 = AC2 [By Pythagoras theorem]
=> BC2 = AC2 + AB2 + AB × AQ + AB × AQ
=> BC2 = AC2 + AB2 + AB × AQ + AP × AC         [From (1)]
=>BC2 = AC2 + AP × AC + AB2 + AB × AQ
=> BC2 = AC (AC + AP) + AB (AB + AQ)
=> BC2 = AC × CP + AB × BQ 

Hope it helps!

Answer 2

Refer to the attachment.