MathDB

2018.11.2

Part of Kyiv City MO Seniors Round2 2010+ geometry

Problems(1)

right angle wanted, AB=BC, circumcircle related (2018 Kyiv City MO Round2 11.2)

Source:

9/14/2020
In the quadrilateral ABCDABCD , AB=BCAB = BC , the point KK is the midpoint of the side CDCD , the rays BKBK and ADAD intersect at the point MM , the circumscribed circle ΔABM \Delta ABM intersects the line ACAC for the second time at the point PP . Prove that ∠BKP=90∘\angle BKP = 90 {} ^ \circ .
(Anton Trygub)
circumcirclegeometryright angle