The bisectors of angles C and D of trapezoid ABCD meet at point P, which lies on side AB.
| February 22, 2021 | education
The bisectors of angles C and D of trapezoid ABCD meet at point P, which lies on side AB. Prove that point P is equidistant from lines BC, CD and AD.
Since point P lies on the bisector of angle C, it means that it is equidistant from the sides of angle BC and CD (any point of the bisector of an unfolded angle is equidistant from the sides of this angle).
Similarly, since point P lies on the bisector of angle D, it is equidistant from the sides of this angle CD and AD.
Therefore, point P is equidistant from all three lines BC, CD and AD.
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