## DES A 38 en given figure LABC = LDA AB = 8cm, AC=4cm, AD=5cm Prove that ACD is similar to BCA and find B

Question

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## Answers ( )

Answer:AB = 8 cm,

AC = 4 cm, AD = 5 cm.

(i) In ΔACD and ΔBCA

∠ABC = ∠DAC (Given)

∠ACD = ∠BCA (Common)

⇒ ΔACD ~ ΔBCA (AA axiom).

Hence ΔACD is similar to ΔBCA.

Hence proved.

(ii) As we have,

AC/BC = CD/CA = AD/BA

⇒ 4/BC = CD/4 = 5/8

⇒ 4/BC = 5/8

⇒ BC = 8 × 4/5 = 32/5

= 6.4 cm.

And CD/4 = 5/8

⇒ CD = 5 × 4/8

⇒ CD = 2.5 cm.

(iii) Area of ΔACD/Area of ΔABC = (AC/AB)2

= (4/8)2

= ¼

Thus area of ΔACD : area of ΔABC = 1 : 4.Step-by-step explanation: