Question

In the given figure, altitudes AD and CE of ∆ ABC intersect each other at the point P. Show that:

(i) ∆AEP ~ ∆ CDP

(ii) ∆ABD ~ ∆ CBE

(iii) ∆AEP ~ ∆ADB

(iv) ∆ PDC ~ ∆ BEC

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Answer 1

(i) In ∆AEP and ∆CDP, we have

∠AEP = ∠CDP [each equal to 90°]

∠APE = ∠CPD [vertically opposite angles]

∴ ∆AEP ~ ∆CDP [by AA-similarity]

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(ii) In AABD and ACBE, we have

∠ADB = ∠CEB = 90°

∠B = ∠B [common]

∴ ∆ABD ~ ∆CBE [by AA-similarity]

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(iii) In ∆AEP and ∆ADB, we have

∠AEP = ∠ADB = 90°

∠EAP = ∠DAB [common]

Hence, ∆AEP ~ ∆ADB [by AA-similarity]

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(iv) In ∆PDC and ∆BEC, we have

∠PDC = ∠BEC = 90°

∠PCD = ∠BCE. [common]

∴ ∆PDC ~ ∆BEC [by AA-similarity]