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# In the figure, altitudes AD and CE of $$\triangle$$ ABC intersect each other at the point P. Show that: (i) $$\triangle$$ AEP ~ $$\triangle$$ CDP (ii) $$\triangle$$ ABD ~ $$\triangle$$ CBE (iii) $$\triangle$$ AEP ~ $$\triangle$$ ADB (iv) $$\triangle$$ PDC ~ $$\triangle$$ BEC.

Given, altitudes AD and CE of $$\triangle$$ ABC intersect each other at the point P.
(i) In $$\triangle$$ AEP and $$\triangle$$ CDP,
$$\angle$$ AEP = $$\angle$$ CDP (90° each)
$$\angle$$ APE = $$\angle$$ CPD (Vertically opposite angles)
Hence, by AA similarity criterion,
$$\triangle$$ AEP ~ $$\triangle$$ CDP

(ii) In $$\triangle$$ ABD and $$\triangle$$ CBE,
$$\angle$$ ADB = $$\angle$$ CEB ( 90° each)
$$\angle$$ ABD = $$\angle$$ CBE (Common Angles)
Hence, by AA similarity criterion,
$$\triangle$$ ABD ~ $$\triangle$$ CBE

(iii) In $$\triangle$$ AEP and $$\triangle$$ ADB,
$$\angle$$ AEP = $$\angle$$ ADB (90° each)
$$\angle$$ PAE = $$\angle$$ DAB (Common Angles)
Hence, by AA similarity criterion,
$$\triangle$$ AEP ~ $$\triangle$$ ADB

(iv) In $$\triangle$$ PDC and $$\triangle$$ BEC,
$$\angle$$ PDC = $$\angle$$ BEC (90° each)
$$\angle$$ PCD = $$\angle$$ BCE (Common angles)
Hence, by AA similarity criterion,
$$\triangle$$ PDC ~ $$\triangle$$ BEC
Hence, by AA similarity criterion,
$$\triangle$$ PDC ~ $$\triangle$$ BEC