An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the altitude from A to BC passes through the centre of the circle.
Let the altitude from A to BC intersect BC at M. Since it is an altitude, AM is perpendicular to BC, meaning ∠AMB = ∠AMC = 90°.
In ΔABM and ΔACM, we have AB = AC (given), ∠AMB = ∠AMC = 90°, and AM = AM (common side). By RHS congruence rule, ΔABM ≅ ΔACM. Therefore, BM = CM, which means M is the midpoint of BC.
Since AM is perpendicular to chord BC and passes through its midpoint M, AM is the perpendicular bisector of chord BC. The centre of the circle lies on the perpendicular bisector of any chord. Hence, the altitude AM must pass through the centre of the circle.
Explanation
The question requires proving that the altitude from the apex of an isosceles triangle inscribed in a circle passes through the circle's centre. The textbook context establishes that the centre of a circle (circumcentre) lies on the perpendicular bisectors of the triangle's sides (chords). By proving the altitude from A bisects the base BC (making it the perpendicular bisector), we can conclude it passes through the centre. The hint in the context likely contains a typo ('show AM = BM' instead of 'BM = CM'), so the standard logical flow of proving the altitude is the perpendicular bisector is used.
Solution Steps
Step 1: Let the altitude from A meet BC at M. Thus, AM ⊥ BC, so ∠AMB = ∠AMC = 90°.
Step 2: Prove M is the midpoint of BC. In ΔABM and ΔACM, AB = AC (given), ∠AMB = ∠AMC = 90°, and AM = AM (common). By RHS congruence, ΔABM ≅ ΔACM, so BM = CM.
Step 3: Since AM is perpendicular to BC and M is its midpoint, AM is the perpendicular bisector of chord BC.
Step 4: Because the centre of the circle lies on the perpendicular bisector of any chord, the altitude AM passes through the centre.