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In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: (iii) ΔAEP ~ ΔADB Q.7(3)

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How i solve the question of class 10th math of Triangles chapter of exercise 6.3 in easy way, because it is very important so please guide me the best way to solve this question In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: (iii) ΔAEP ~ ΔADB

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  1. In ΔAEP and ΔADB,

    ∠AEP = ∠ADB (90° each)

    ∠PAE = ∠DAB (Common Angles)

    Hence, by AA similarity criterion,

    ΔAEP ~ ΔADB

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