In Δ ABC, ∠A is obtuse, PB⊥AC and QC⊥AB. Prove that:

(i) AB×AQ=AC×AP

(ii) BC² =(AC×CP+AB×BQ)

Given: PB⊥AC and QC⊥AB

To prove: (i) AB×AQ=AC×AP

(ii) BC² =(AC×CP+AB×BQ)

Theorem Used:

Pythagoras theorem:

In a right-angled triangle, the squares of the hypotenuse is equal to the sum of the squares of the other two sides.

Proof:

i) In ΔAPB and ΔAQC
∠P = ∠Q = 90°
∠PAB = ∠QAC [Vertically opposite angles]

ΔAPB~Δ AQC [By AA similarity]

{Corresponding part of similar triangle are proportional}

AP x AC=AQ x AB ………….(i)

ii) In ΔBPC

BC²=BP²+PC²

BC²=AB²-AP²+(AP+AC)²

BC²=AB²-AP²+AP²+AC²+2APxAC

BC²=AB²+AC²+2APxAC ……..(ii)

In ΔBQC

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=AC x PC + AB x BQ