CBSE Class 10 Answered
ABD is a right angled triangle at A and AC_|_BD show that
(I) AB²=BC.BD
(II) AC²=BC.DC
(III) AD²=BD.CD
Asked by mulkallaswapna | 15 Jan, 2020, 07:32: AM
Expert Answer
1.
ΔADB and ΔCAB
∠DAB = ∠ ACB (Each 90°)
∠ADB = ∠CBA (Common angle)
∴ ΔADB ∼ ΔCAB
AB/CB = BD/AB
AB² = BD*CB
∠DAB = ∠ ACB (Each 90°)
∠ADB = ∠CBA (Common angle)
∴ ΔADB ∼ ΔCAB
AB/CB = BD/AB
AB² = BD*CB
2.
ΔCBA
∠CAB = x
∠CBA = 180° - 90° - x
∠CBA = 90° - x
simillarly,
∠CAD = 90° - x
∠CDA = 180° - 90° - ( 90° - x )
∠CDA = x
In ΔCBA AND ΔCAD
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA
∴ ΔCBA ∼ ΔCAD
AC/DC = BC/AC
AC² = BC * DC
∠CAB = x
∠CBA = 180° - 90° - x
∠CBA = 90° - x
simillarly,
∠CAD = 90° - x
∠CDA = 180° - 90° - ( 90° - x )
∠CDA = x
In ΔCBA AND ΔCAD
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA
∴ ΔCBA ∼ ΔCAD
AC/DC = BC/AC
AC² = BC * DC
3.
ΔDCA and ΔDAB
∠DCA = ∠DAB
∠CDA = ∠ADB
∴ ΔDCA ∼ ΔDAB
DC/DA = DA/DB
AD² = CD * BD
∠DCA = ∠DAB
∠CDA = ∠ADB
∴ ΔDCA ∼ ΔDAB
DC/DA = DA/DB
AD² = CD * BD
Answered by Arun | 19 Jan, 2020, 05:06: PM
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