Question

$\Delta ABD$ is a right triangle right-angled at A and AC \perp BD. Show that
(i) $A{B}^2=BC.BD$
(ii) $A{C}^2=BC.DC$
(iii) $A{D}^2=BD.CD$
(iv) $\frac{A{B}^2}{A{C}^2}=\frac{BD}{DC}$

Answer 1

(i) In ΔADB and ΔCAB, we have
∠DAB = ∠ACB (Each equals to 90°)
∠ABD = ∠CBA (Common angle)
∴ ΔADB ~ ΔCAB [AA similarity criterion]


(ii) Let ∠CAB = x
In ΔCBA,
∠CBA = 180° - 90° - x
∠CBA = 90° - x
Similarly, in ΔCAD
∠CAD = 90° - ∠CBA
= 90° - x
∠CDA = 180° - 90° - (90° - x)
∠CDA = x
In ΔCBA and ΔCAD, we have
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA (Each equals to 90°)
∴ ΔCBA ~ ΔCAD [By AAA similarity criterion]


(iii) In ΔDCA and ΔDAB, we have
∠DCA = ∠DAB (Each equals to 90°)
∠CDA = ∠ADB (common angle)
∴ ΔDCA ~ ΔDAB [By AA similarity criterion]
⇒
=>

(iv)