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In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:(iv) ΔPDC ~ ΔBEC Q.7(4)

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The last question of qo.7(4) of triangles chapter of exercise 6.3, how i solve the question of class 10th math of triangles chapter please give me the best solution of this question  In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:(iv) ΔPDC ~ ΔBEC

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  1. In ΔPDC and ΔBEC,

    ∠PDC = ∠BEC (90° each)

    ∠PCD = ∠BCE (Common angles)

    Hence, by AA similarity criterion,

    ΔPDC ~ ΔBEC

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