⊿APB~⊿AQC [By AA similarity]

AP/AQ=AQ/AB {Corresponding part of similar triangle are proportional}

(II) APxAC=AQxAB ………….(i)

In ⊿BPC

BC²=BP²+PC²

BC²=AB²-AP²+(AP+AC)²

BC²=AB²-AP²+AP²+AC²+2APxAC

BC²=AB²+AC²+2APxAC ……..(ii)

In ⊿BQC

BC²=CQ²+BQ²

BC²=AC²-AQ²+(AB+AQ)²

BC²=AC²-AQ²+AB² +2ABxAQ

BC²=AC² +AB²+AQ²+2ABxAQ ………….(iii)

Adding equ. (ii)and(iii)

2BC²=2AC²+2AB²+2APxAC+2ABxAQ

2BC²=2AC[AC+AP]+AB[AB+AQ]

2BC²=2ACxPC+2ABxBQ

BC²=ACxPC+ABxBQ