In the adjoining figure, $A B = A C$ and $A P ⊥ B C$ . Then:

A B = A P

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A B < A P

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A B > A P

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A B \leq A P

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Solution

The correct option is C $A B > A P$
Given: $A B = A C$ , $A P ⊥ B C$
In $△ A B P$ , we have
$∠ A P B = 90^{\circ}$
$∠ A B P = 90 - ∠ B A P$ ....(Angle sum property)
Thus, $∠ A P B > ∠ A B P$
$A B > A P$ .... (Side opposite to larger angle is large)

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