Question

in circle o, chords bf and ce are perpendicular to diameter ad. which of the following relationships is true? ag = oh ch = he bg = ch bg = he

Explanation:

Step1: Recall the perpendicular - chord theorem

If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord.
In circle \(O\), diameter \(AD\) is perpendicular to chord \(CE\). Let the intersection point of \(AD\) and \(CE\) be \(H\).
By the perpendicular - chord theorem, if a line (in this case, the diameter \(AD\)) is perpendicular to a chord (\(CE\)) of a circle, then it bisects the chord. So, \(CH=\frac{1}{2}CE\) and \(HE = \frac{1}{2}CE\).

Step2: Conclude the relationship

Since \(CH=\frac{1}{2}CE\) and \(HE=\frac{1}{2}CE\), we have \(CH = HE\).

For the other options:

• \(AG = OH\): There is no property that directly relates \(AG\) and \(OH\) based on the given information (perpendicular chords to a diameter).
• \(BG = CH\): There is no guarantee that chords \(BF\) and \(CE\) are congruent, so we cannot say \(BG = CH\).
• \(BG=HE\): Similarly, there is no guarantee that chords \(BF\) and \(CE\) are congruent, so we cannot say \(BG = HE\).

Answer:

CH = HE