Bisectors of angles A and B of trapezoid ABCD intersect at point K lying on the side of the CD.

| February 10, 2021 | education

Bisectors of angles A and B of trapezoid ABCD intersect at point K lying on the side of the CD. prove that point K is equidistant from straight lines AB, BC and AD.

The distance from a point to a straight line is measured along the perpendicular drawn from a point to a given straight line. Let us drop the perpendiculars from point K to the straight lines AD, AB and BC. We get the segments KM, KP and KН, respectively. Consider right-angled triangles KMA and KРA, they have a common hypotenuse and equal acute angles (AK – bisector), so the triangles are equal, hence KM = KP. Now let us consider the right-angled triangles KНВ and KРВ. Here is exactly the same picture – the general hypotenuse and equal acute angles (ВK – bisector), which means KH = KR. KM = KР = KН, point K is equidistant from the sides of AD, AB and BC.



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