In \(\triangle ABC\), AB = AC and O is the mid-point of BC. OD is perpendicular to AB and OE is perpendicular to AC. If \(\angle BOD=40^\circ\), what is the measure of \(\angle AOE\)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 50°
In an isosceles triangle the line from the apex to the mid-point of the base is perpendicular to the base, so \(\angle AOC=90^\circ\). The equal base angles then give \(\angle COE\).
In \(\triangle BOD\): \(\angle B=180^\circ-90^\circ-40^\circ=50^\circ\)
AB = AC, so \(\angle C=\angle B=50^\circ\)
In \(\triangle COE\): \(\angle COE=180^\circ-90^\circ-50^\circ=40^\circ\)
\(\angle AOE=\angle AOC-\angle COE=90^\circ-40^\circ=50^\circ\)
