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In following figure. altitudes AD and CE of ∆ABC Intersect each other at the point P. Show that:(i) ∆AEP ~ ∆CDP (ii) ∆ABD ~ ∆CBE (iii) ∆AEP ~ ∆AÐB (iv) ∆PDC ~ ∆BEC. |
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Answer» Data: altitudes AD and CE of ABC Intersect each other at the point P. To Prove: (i) ∆AEP ~ ∆CDP (ii) ∆ABD ~ ∆CBE (iii) ∆AEP ~ ∆ADB (iv) ∆PDC ~ ∆BEC (i) In ∆AEP and ∆CDP. ∠AEP = ∠CDP = 90° (data) ∠APE = ∠CPD (Vertically opposite angle) ∴ ∠PAE = ∠PCD These are equiangular triangles. Similarity criterion for ∆ is A.A.A ∴ ∆AEP ~ ∆CDP (ii) In ∆ABD and ∆CBE. ∠ADB = ∠CEB = 90° (data) ∠ABD = ∠CBE (common) ∴ ∠DAB = ∠BCE These are equiangular triangles. ∴ Similarity criterion for is A.A.A. ∴ ∆ABD ~ ∆CBE (iii) In ∆AEP and ∆ADB. ∠AEB = ∠ADB = 90° (data) ∠PAE = ∠DAB (common) ∴ ∠APE = ∠ABD. ∴ These are equiangular triangles. ∴ Similarity criterion for is A.A.A. ∴ ∆AEP ~ ∆ADB (iv) In ∆PDC and ∆BEC. ∠PDC = ∠BEC = 90° (data) ∠PCD = ∠BCE (common) ∴ ∠CPD = ∠CBE ∴ These are equiangular angular triangles. Similarity criterion for ∆ is A.A.A. ∴ ∆PDC ~ ∆BEC. |
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