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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 2 of 4: Deduce equal distances
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In plain words

Deduce equal distances

∠OEQ=∠OBQ=∠OCQ=∠OFQ\angle OEQ=\angle OBQ=\angle OCQ=\angle OFQ
Detailed analysis

From the first cyclic quadrilateral, ∠OEQ=∠OBQ\angle OEQ=\angle OBQ; from the second, ∠OFQ=∠OCQ\angle OFQ=\angle OCQ. Since AB=ACAB=AC and OO lies on the symmetry axis AMAM, ∠OBQ=∠OCQ\angle OBQ=\angle OCQ. Therefore ∠OEQ=∠OFQ\angle OEQ=\angle OFQ. Both triangles OEQOEQ and OFQOFQ have right angle at QQ and common hypotenuse-side data, so they are congruent and QE=QFQE=QF.