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In fig. 6.38 altitudes AD and CE intersect each other at the point P. Show that |
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Answer» AD and CE are altitudes, which intersect each other at P.(i) In ∆AEP and ∆CDP∠AEP = ∠CDP =\xa090°\xa0[given]and ∠APE = ∠CPD[vertically opposite angles]Therefore, by using AA similar condition∆AEP ~ ∆CDP.(ii) In ∆ABD and ∆CBE∠ADB = ∠CEB = 90° [given]and ∠B = ∠B [common]Therefore, by using AA similar condition∆ABD ~ ∆CBE.(iii) In ∆AEP and ∆ADB∠AEP = ∠ADB = 90° [given]and ∠PAE = ∠DAB [common]Therefore, by using AA similar condition∆AEP ~ ∆ADB(iv) In ∆PDC and ∆BEC∠PDC = ∠CEB = 90° [given]∠PCD = ∠ECB [common]Therefore, by using AA similar condition∆PDC ~ ∆BEC. In the given Fig,\xa0altitudes AD and CE of\xa0\xa0intersects each other at the point P. Show that:(i) ∆AEP ~ ∆CDP(ii) ∆ABD ~ ∆CBE(iii) ∆AEP ~ ∆ADB(iv) ∆PDC ~ ∆BEC. \xa0AD and CE are altitudes, which intersect each other at P.(i) In ∆AEP and ∆CDP∠AEP = ∠CDP =\xa090°\xa0[given]and ∠APE = ∠CPD[vertically opposite angles]Therefore, by using AA similar condition∆AEP ~ ∆CDP.(ii) In ∆ABD and ∆CBE∠ADB = ∠CEB = 90° [given]and ∠B = ∠B [common]Therefore, by using AA similar condition∆ABD ~ ∆CBE.(iii) In ∆AEP and ∆ADB∠AEP = ∠ADB = 90° [given]and ∠PAE = ∠DAB [common]Therefore, by using AA similar condition∆AEP ~ ∆ADB(iv) In ∆PDC and ∆BEC∠PDC = ∠CEB = 90° [given]∠PCD = ∠ECB [common]Therefore, by using AA similar condition∆PDC ~ ∆BEC. |
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