In the given figure, ∠ ADC = ∠ BCA, prove that Δ ACB ~ Δ ADC

Q) In the given figure, ∠ ADC = ∠ BCA, prove that Δ ACB ~ Δ ADC. Hence find if AC = 8 cm and AD = 3 cm.

Ans:

Ans: In Δ ACB and Δ ADC,

∠ ADC = ∠ BCA (given)

∠ A = ∠ A (common angle)

$\therefore$ Δ ACB ~ Δ ADC (by AA similarity identity)

Hence proved.

Next, from above similar triangles, we get $\frac{AC}{AD} = \frac{AB}{AC}$

Hence, $\frac{8}{3} = \frac{AB}{8}$

AB = $\frac{64}{3}$

BD = AB – AD = $\frac{64}{3}$ – 3 = $\frac{55}{3}$

Therefore, BD = 55/3 cm.

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