MathLabs

Problem 2

In isosceles triangle ABCABC with AB=ACAB=AC, let MM be the midpoint of BCBC and let O∈AMO\in AM satisfy OB⊥ABOB\perp AB. For Q∈BCQ\in BC, let E∈ABE\in AB and F∈ACF\in AC be distinct collinear points with QQ. Prove OQ⊥EFOQ\perp EF if and only if QE=QFQE=QF.
Step 4 of 4: Finish by contradiction
Previous stepNext step
In plain words

Finish by contradiction

△QEE′≅△QFF′⇒AB∥AC\triangle QEE'\cong\triangle QFF'\Rightarrow AB\parallel AC
Detailed analysis

Because E,Q,FE,Q,F and E′,Q,F′E',Q,F' are straight and QE=QFQE=QF, QE′=QF′QE'=QF', the triangles QEE′QEE' and QFF′QFF' are congruent by SAS (the included angles at QQ are equal). Thus ∠QEE′=∠QFF′\angle QEE'=\angle QFF'. But EE′⊂ABEE'\subset AB and FF′⊂ACFF'\subset AC, while EQEQ and FQFQ lie on the same line in opposite directions; these equal angles imply AB∥ACAB\parallel AC, impossible for a nondegenerate triangle. Hence the original EFEF must be perpendicular to OQOQ.