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# In Figure, ABD is a triangle right angled at A and AC $$\perp$$ BD. Show that (i) $$AB^2 = BC × BD$$ (ii) $$AC^2 = BC × DC$$ (iii) $$AD^2 = BD × CD$$

(i) In $$\triangle$$ ADB and $$\triangle$$ CAB,
$$\angle$$ DAB = $$\angle$$ ACB (Each 90°)
$$\angle$$ ABD = $$\angle$$ CBA (Common angles)
$$\triangle$$ ADB ~ $$\triangle$$ CAB [AA similarity criterion]
AB/CB = BD/AB
$$AB^2 = CB × BD$$

(ii) Let $$\angle$$ CAB = x
In $$\triangle$$ CBA,
$$\angle$$ CBA = 180° – 90° – x
$$\angle$$ CBA = 90° – x

Similarly, in $$\triangle$$ CAD
$$\angle$$ CAD = 90° – $$\angle$$ CBA
= 90° – x
$$\angle$$ CDA = 180° – 90° – (90° – x)
$$\angle$$ CDA = x

In $$\triangle$$ CBA and $$\triangle$$ CAD, we have
$$\angle$$ CBA = $$\angle$$ CAD
$$\angle$$ CAB = $$\angle$$ CDA
$$\angle$$ ACB = $$\angle$$ DCA (Each 90°)
$$\triangle$$ CBA ~ $$\triangle$$ CAD [AAA similarity criterion]
AC/DC = BC/AC
$$AC^2 = DC × BC$$

(iii) In $$\triangle$$ DCA and $$\triangle$$ DAB,
$$\angle$$ DCA = $$\angle$$ DAB (Each 90°)
$$\angle$$ CDA = $$\angle$$ ADB (common angles)
$$\triangle$$ DCA ~ $$\triangle$$ DAB [AA similarity criterion]
DC/DA = DA/DA
$$AD^2 = BD × CD$$