Summary of Triangle Theorems
- Centroid: intersection of the medians, AG:GM=2:1
- Circumcenter: intersection of the perpendicular bisectors, circumcenter R
- Incenter: intersection of the angle bisectors, inradius r=S/s
- Orthocenter: intersection of the altitudes
- Ceva: BD/DC×CE/EA×AF/FB=1 ⟺ the three lines intersect at one point
- Menelaus: BD/DC×CE/EA×AF/FB=1 ⟺ the three points lie on one straight line
- Inscribed angle=central angle/2, sum of opposite angles in a cyclic quadrilateral=180°
- Power of a point: TP×TQ=TR×TS (intersection・external point), TP²=TQ×TR (tangent)
Comprehensive Practice Problems
- For a right triangle with side lengths 6, 8, and 10, find (a) the inradius and (b) the circumradius.
- In △ABC, BD:DC=2:3 and CE:EA=1:4. Let P be the intersection of AD and BE, and let the extension of CP intersect AB at F. Find AF:FB (use Ceva's theorem).
- From external point T, draw a tangent touching the circle at P (TP=6). A secant from T intersects the circle at Q and R (TQ＜TR). If TQ=3, find TR.
Solutions and Explanations
- (a) S=(1/2)×6×8=24、s=(6+8+10)/2=12 → r=24/12=2
(b) Circumradius of a right triangle=hypotenuse/2=10/2=5 - By Ceva's theorem, (BD/DC)×(CE/EA)×(AF/FB)=1
(2/3)×(1/4)×(AF/FB)=1 → AF/FB=6 → AF:FB=6:1 - TP²=TQ×TR → 36=3×TR → TR=12