Plane geometry II · Chapter III · Consolidate

Problem 3, 2018

← Prev · 28 / 35 · Next →

RegionalProof

The circle inscribed in triangle \(ABC\) touches the sides \(BC\), \(CA\) and \(AB\) at the points \(D\), \(E\) and \(F\) respectively. A point \(K\) lies on the same side of the line \(EF\) as the vertex \(A\), and satisfies

\[ \angle KFE = \angle ACB \qquad \text{and} \qquad \angle KEF = \angle ABC . \]

Prove that \(KD \perp BC\).

A B C D E F K
The given configuration. The angle of triangle \(KEF\) at \(F\) equals the angle of \(ABC\) at \(C\), and the angle at \(E\) equals the angle at \(B\).

Sign in to check answers, open hints, read the full solution, and track your progress. Statements are always free.

Serbian Regional Competition 2018, high school grade I, category A, problem 3. Organized by the Mathematical Society of Serbia (DMS). Source