MathDB

2015.11.4.1

Part of Kyiv City MO Seniors 2003+ geometry

Problems(1)

MB = MC wanted, angle bisector, perpendiculars (2015 Kyiv City MO 11.4.1)

Source:

9/3/2020
On the bisector of the angle BAC BAC of the triangle ABC ABC we choose the points B1,  C1 {{B} _ {1}}, \, \, {{C} _ {1}} for which BB1⊥AB B {{B} _ {1 }}\perp AB , CC1⊥AC C {{C} _ {1}} \perp AC . The point M M is the midpoint of the segment B1C1 {{B} _ {1}} {{C} _ {1}} . Prove that MB=MC MB = MC .
geometryangle bisectorequal segmentsperpendicular