MathDB

2011.8.3

Part of Kyiv City MO Juniors Round2 2010+ geometry

Problems(1)

right angle wanted inside a square, AM = BN (2011 Kyiv City MO Round2 8.3)

Source:

7/31/2020
On the sides AD,BCAD , BC of the square ABCDABCD the points M,NM, N are selected NN, respectively, such that AM=BNAM = BN. Point XX is the foot of the perpendicular from point DD on the line ANAN. Prove that the angle MXCMXC is right.
(Mirchev Borislav)
geometryright anglesquareequal segments