Question

In ∆ABC, AP ⊥ BC, BQ ⊥ AC B– P–C, A–Q – C then prove that, ∆CPA ~ ∆CQB. If AP = 7, BQ = 8, BC = 12 then Find AC.

Answer 1

Given:
AP ⊥ BC
BQ ⊥ AC
To prove: ∆CPA ~ ∆CQB
Proof: In ∆CPA and ∆CQB
∠CPA = ∠CQB = 90^∘ (Given)
∠C = ∠C (Common)
By AA test of similarity
∆CPA ~ ∆CQB
Hence proved.
$$\begin{gathered} \mathrm{Now},\frac{\mathrm{AP}}{\mathrm{BQ}}=\frac{{\displaystyle \mathrm{AC}}}{{\displaystyle \mathrm{BC}}}\left(\mathrm{Corresponding}\mathrm{sides}\mathrm{are}\mathrm{proportional}\right) \\ \Rightarrow \mathrm{AC}=\frac{\mathrm{AP}}{\mathrm{BQ}}\times \mathrm{BC} \\ =\frac{7}{8}\times 12 \\ =10.5 \end{gathered}$$​