Question

In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.

1. Prove that ΔACD is similar to ΔBCA.
2. Find BC and CD.
3. Find the area of ΔACD : area of ΔABC.

Answer 1

∠ABC = ∠DAC = x  ...(Say)

AB = 8 cm,

AC = 4 cm,

AD = 5 cm.

(i)

In ΔACD and ΔBCA

∠ABC = ∠DAC   ...(Given)

∠ACD = ∠BCA   ...(Common angles)

`=>` ΔACD ∼ ΔBCA   ...(AA criterion for similarity) (i)

Hence, ΔACD is similar to ΔBCA.

(ii)

In ΔACD and ΔBCA 

ΔACD ∼ ΔBCA   ...[From (i)]

`(AC)/(BC) = (CD)/(CA) = (AD)/(BA)`

`(4)/(BC) = (CD)/(4) = (5)/(8)`

`(4)/(BC) = (5)/(8)`

`BC = (8 xx 4)/(5) = (32)/(5)`

∴ BC = 6.4 cm

And `(CD)/(4) = (5)/(8)`

`CD = (5 xx 4)/(8)`

∴ CD = 2.5 cm

(iii)

In ΔACD and ΔBCA 

ΔACD ∼ ΔBCA   ...[From (i)] 

`"area of ΔACD"/"area of ΔABC" = ("AD"/"AB")^2`

`"area of ΔACD"/"area of ΔABC" = (5/8)^2`

= `25/64`

∴ area of (ΔACD) : area of (ΔABC) = 25 : 16