Question

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

(i) AB x AQ = AC x AP

(ii) \(BC^2 = \)(AC x CP + AB x BQ)

Answer 1

Draw the diagram according to given questions.

(I) In ⊿APB and ⊿AQC∠A = ∠A

(common)∠P = ∠Q ( both 90°)

∴ ⊿APB~⊿AQC [By AA similarity]

⇒ \(\frac{AP}{AQ}\) = \(\frac{AB}{AC}\) {Corresponding part of similar triangle are proportional}

AP x AC = AQ x AB ………….(1)

(II) In ⊿BPC By pythagoras theoram, 

BC² = BP² + PC² Also in ⊿BPA 

BP² = AB² - AP² Also PC = PA + AC 

⇒ BC² = AB² - AP² + (AP + AC)²  

Apply the theorem (a + b)² = a² + b² + 2ab in (AP + AC)² 

⇒ BC² = AB² - AP² + AP² + AC² + 2AP x AC 

BC² = AB² + AC² + 2AP x AC ……..(ii) 

In ⊿BQC 

BC² = CQ² + BQ² 

BC2 = AC2 - AQ2 + (AB + AQ)2 

BC² = AC² - AQ² + AB² + 2AB x AQ 

BC² = AC² + AB² + AQ² + 2AB x AQ ………….(iii) 

Adding equ. (ii)and(iii)

BC² + BC² = AB² + AC² + 2AP x AC + AC² + AB² + AQ² + 2AB x AQ

⇒ 2BC² = 2AC² + 2AB² + 2AP x AC + 2AB x AQ 

⇒ 2BC² = 2AC[AC + AP] + AB[AB + AQ] 

⇒ 2BC² = 2AC x PC + 2AB x BQ 

⇒ BC² = AC x PC + AB x BQ

Hence proved.