In triangle ABC , AC is perpendicular to BC and CD perpendicular to AB Prove that BC2/ AC2 = - Maths - Triangles

Question

In triangle ABC , AC is perpendicular to BC and CD perpendicular
to AB Prove that BC2/ AC2 = BD/AD

Harleen Birdi · Accepted Answer

Consider Î”ACD and Î”ABC, âˆ CAD = âˆ CAB, âˆ CDA = âˆ ACB $∴ Δ ACD \sim Δ ABC, by AA similarity criterion \Rightarrow \frac{AC}{AB} = \frac{AD}{AC} \Rightarrow {AC}^{2} = AB \times AD (1) Similarly, Δ BCD \sim Δ BAC (by AA similarity criterion) \Rightarrow \frac{BC}{BA} = \frac{BD}{BC} \Rightarrow {BC}^{2} = BA \times BD (2) On dividing (2) by (1), we get \frac{{BC}^{2}}{{AC}^{2}} = \frac{AB \times BD}{AB \times AD} \Rightarrow \frac{{BC}^{2}}{{AC}^{2}} = \frac{BD}{AD}$

In triangle ABC , AC is perpendicular to BC and CD perpendicular to AB. Prove that BC2/ AC2 = BD/AD

In triangle ABC , AC is perpendicular to BC and CD perpendicular to AB. Prove that BC²/ AC² = BD/AD