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In fig. 6.38 altitudes AD and …

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In fig. 6.38 altitudes AD and CE intersect each other at the point P. Show that

    • Posted by Sandhya Chahar 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Ganesh G 5 years, 1 month ago

    AD and CE are altitudes, which intersect each other at P. (i) In ∆AEP and ∆CDP ∠AEP = ∠CDP = 90° [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.

    0Thank You

    Gaurav Seth 5 years, 1 month ago

    In the given Fig, altitudes AD and CE of  intersects each other at the point P. Show that:
    (i) ∆AEP ~ ∆CDP
    (ii) ∆ABD ~ ∆CBE
    (iii) ∆AEP ~ ∆ADB
    (iv) ∆PDC ~ ∆BEC.

    <hr />


     

    AD and CE are altitudes, which intersect each other at P.
    (i) In ∆AEP and ∆CDP
    ∠AEP = ∠CDP = 90° [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.

    0Thank You

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