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Question
In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.
- Prove that ΔACD is similar to ΔBCA.
- Find BC and CD.
- Find area of ΔACD : area of ΔABC.
Solution
∠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)`
= 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" = ("AC"/"AB")^2`
`=> "Area of ΔACD"/"Area of ΔABC" = 4^2/8^2 = 16/64 = 1/4`
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