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प्रश्न
In Figure ABD is a triangle right angled at A and AC ⊥ BD. Show that AC2 = BC × DC
उत्तर
Let ∠CAB = x
In ΔCBA,
∠CBA = 180° - 90° - x
∠CBA = 90° - x
Similarly, in ΔCAD
∠CAD = 90° - ∠CBA
= 90° - x
∠CDA = 180° - 90° - (90° - x)
∠CDA = x
In ΔCBA and ΔCAD, we have
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA (Each equals to 90°)
∴ ΔCBA ~ ΔCAD [By AAA similarity criterion]
`⇒ (AC)/(DC) = (BC)/(AC)`
⇒ AC2 = DC × BC
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