Point P is the incenter of DEF. Point P is the point of concurrency of the angle bisector. Find AP.

5

Step-by-step explanation:

because line PF DIVIDED LINE BPA INTO TWO EQUAL HALF THEREFORE LINE BP IS EQUAL TO LINE AB WHICH IS EQUAL TO 5

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.

the answer is c.) yes, [triangle]klp can be reflected across the line containing [line]kp and then translated so that p is mapped to m.

the reason is this:

looking at the triangles visually we can see they share the same features. using rigid transfromations we can reflect it over [line]kp then all we need to do is translate up.