P is a point equidistant from two lines l and m intersecting at point A. Show that the line AP bisects that angle between them
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Solution
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You are given that lines l and m intersect each other at A.
Let PB is perpendicular on l and
PC⊥m. It is given that PB=PC.
You need to show that ∠PAB=∠PAC.
Let us consider △PAB and △PAC. In these two triangles,
PB=PC (Given)
∠PBA=∠PCA=90∘ (Given)
PA=PA (Common side)
So, △PAB≅△PAC (RHS rule)
So, ∠PAB=∠PAC(CPCT)
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