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.
Step-by-step solution
To find: Show that O lies on AD
Idea: Use the locus fact: the points equidistant from B and C are exactly the points on the perpendicular bisector of BC. Show A and O are both such points, then notice that there is only one perpendicular from A to BC.
- OB = OC (radii), so O is equidistant from B and C; hence O lies on the perpendicular bisector l of BC.1 mark
- AB = AC (given), so A is also equidistant from B and C; hence A lies on l.½ mark
- So l is a line through A that is perpendicular to BC. From a point there is only one perpendicular to a given line, so l is the altitude AD from A.1 mark
- O lies on l, that is, on the altitude AD. So the altitude from A passes through the centre.½ mark
Answer to write in the exam
Given: AB = AC, ΔABC inscribed in a circle with centre O; AD ⟂ BC. To prove: O lies on AD.
OB = OC (radii) ⇒ O lies on the perpendicular bisector of BC
AB = AC (given) ⇒ A lies on the perpendicular bisector of BC
⇒ The perpendicular bisector of BC passes through A and is ⟂ BC
Only one perpendicular can be drawn from A to BC ⇒ perpendicular bisector of BC = altitude AD
∴ The altitude AD passes through the centre O.
Common mistakes that cost marks
- Assuming the altitude passes through O and then proving something else. The proof must show O is on the altitude.
- Saying “the altitude bisects BC” without a reason. It follows here because the altitude is the perpendicular bisector of BC (A is equidistant from B and C).
- Forgetting that O may lie outside the triangle (when ∠A is obtuse). The proof still works, because it is about the whole line AD, not just the segment inside the triangle.
How this can come in the exam
ΔPQR is inscribed in a circle with centre O, and PQ = PR. PS is the altitude to QR. Which statement is true?
- O lies on PS
- O lies on QR
- O is the midpoint of PQ
- PS = QR
Show answer
(A) O lies on PS
P and O are both equidistant from Q and R, so both lie on the perpendicular bisector of QR, which is the altitude PS.
Try one yourself
In ΔABC, AB = AC = 10 cm and BC = 12 cm. Find the radius of its circumcircle.
Show answer
Altitude AD bisects BC: BD = 6, AD = √(100 − 36) = 8 cm. O is on AD with OB = OA = R, so OD = 8 − R and R2 = 62 + (8 − R)2 ⇒ 16R = 100 ⇒ R = 6.25 cm.
More questions like this
- Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.
- Take a paper circle. Fold the circle from the boundary, inwards. Open the fold. The crease is now a chord (see the figure B). Now fold the paper again, so that the end points of the chord meet. Open the fold (see the figure C).
Measure the lengths of the parts into which the chord is divided. The chord gets bisected where the folds intersect. Measure the angle between the creases. The crease of the second fold is along the perpendicular from the centre to the chord. Measure the distance from the centre to the midpoint of the chord. It is the distance from the centre to the chord.
Now draw another chord of the same length. How will you do this? We will let you figure this out yourself. Join the centre to the midpoint of the new chord and measure its length. Is it the same as distance from the centre to the first chord? - Chords of a circle having the same length are all at the same distance from the centre of the circle.
- Use the Baudhāyana–Pythagoras theorem to show why chords of a circle having the same length are all at the same distance from the centre of the circle.
- Consider the figure. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.