Question

In the given figure, ∠ACB = ∠CDA, AC = 8cm, AD = 3cm, then BD is​

Answer 1

Answer:

Given,

AC = 8 cm, AD = 3cm and ∠ACB = ∠CDA

From figure, ∠CDA = 90°

∠ACB = ∠CDA = 90°

Answer 2

Given:

∠ACB = ∠CDA,

AC = 8cm,

AD = 3cm

From figure , figure are in attachment

$\begin{array}{l}\angle \mathrm{CDA}=90^{\circ} \\\therefore \angle \mathrm{ACB}=\angle \mathrm{CDA}=90^{\circ}\end{array}$

In right angled ∆ADC

$\begin{array}{l}A C^{2}=A D^{2}+C D^{2} \\\\(8)^{2}=(3)^{2}+(C D) \\\\C D^{2}=64-9=55 \\\\C D=\sqrt{55} \mathrm{~cm}\end{array}$

In ∆CDB and ADC

∠BDC = ∠AD [each 90°]

∠DBC = ∠DCA [each equal to 90°-∠A]

∴ ∠CDB ∼ ∆ADC

$\frac{\mathrm{CD}}{\mathrm{D}^{2}}=\frac{\mathrm{AD}}{\mathrm{CD}}[\text { by basic proportionality theorem }]$

$\begin{array}{l}\mathrm{CD}^{2}=\mathrm{AD} \times \mathrm{BD} \\\\\mathrm{BD}=\frac{(\mathrm{CD})^{2}}{\mathrm{AD}}\\\\=\frac{(\sqrt{55})^{2}}{3}\\\\=\frac{55}{3} \mathrm{~cm}\end{array}$

Hence the value of BD is  $\frac{55}{3} \mathrm{~cm}$