Today is our topic of discussion The Circumcenter, Centroid and Orthocenter of a Triangle .
The Circumcenter, Centroid and Orthocenter of a Triangle
Here it should be mentioned that, the distance between the orthocenter and a vertex of a triangle is twice the perpendicular distance from the circumcenter to the opposite side of that vertex.
Circumcenter of a Triangle:
The circumcenter of a triangle is the point of intersection of two perpendicular bisectors of that triangle. Noted that, the perpendicular bisector of the third side of the triangle would pass through the circumcenter too.
Centroid of a Triangle:
The centroid of a triangle is the point of intersection of three medians of that triangle. The centroid of a triangle divides each median into the ratio 2:1.
Orthocenter of a Triangle:
The orthocenter of a triangle is the point of intersection of the perpendiculars drawn from each vertex to their respective opposite side.
Theorem 10.
triangle are collinear. The circumcenter, the centroid and the orthocenter of any
Special Nomination:
Suppose, O is the orthocenter, S’ is the circumcenter and AP is a median of the triangle ΔABC. The line between orthocenter O and the circumcenter S intersect the median AP at point G. If we join S and P, the line SP is a perpendicular on BC. So, it is enough to prove that the point G is the centroid of ΔABC.
Proof:
OA is the distance between the vertex A from the orthocentre O and SP is the distance between the opposite side BC of the vertex A from the circumcentre S of ΔABC. OA=2SP….(1)
Now since AD and SP both are perpendiculars on BC so AD || SP. Now AD || SP and AP is their intersecting line. So ∠PAD LAPS, as they are both alternate angles. Therefore, ∠OAG = ∠SPG.
Now in between ΔAGO and ΔPGS,
∠AGO = ∠PGS [Vertical opposite angle]
∠OAG = ∠SPG [Alternate angles]
remaining ∠AOG = remaining ∠PSG
ΔAGO and ΔPGS equiangular.
So, AG/GP ОА/SP therfore, AG/GP 2SP /SP [from the equation (1)]
Therefore, AG= 2/1 Or, AG: GP = 2:1
So, the point G divides the median AP into the ratio 2: 1.
G is the centroid of ΔABC. [Proved]
Note:
1) Nine Point Circle:
The total nine points including the middle points of three sides, feet of the three perpendiculars drawn from each vertex to the opposite side and the middle points of the three line segments joining the orthocenter to the vertices of any triangle lie on one circle. This circle is called the nine point circle.
2) The center of the nine point circle is the middle point of the line segment joining the orthocenter and the circumcentre.
3) The radius of the nine point circle is half of the circumradius of the triangle.
Theorem 11 (The Theorem of Brahmagupta).
If any cyclic quadrilateral has perpendicular diagonals, then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.
Special Nomination:
ABCD is a quadrilateral inscribed in a circle with perpendicular diagonals AC and BD intersecting at point M. ME is a perpendicular on the side BC from the point M and extended EM intersects the opposite side AD at point F. It is to be proved that AF = FD.
Proof:
Because they are inscribed angles that intercept the same arc CD of the circle, ∠CBD =∠CAD
So, ∠CBM = ∠MAF
Again, ∠CBM = ∠CME [Both complementary to the ∠BME]
So, ∠MAF = ∠FMA
Therefore, in ΔAFM, AF = FM
Similarly it can be shown that, ∠FDM = ∠BCM = ∠BME=∠DMF
So, in ΔDFM, FD = FM
So, AF = FD [Proved]
Theorem 12 (Ptolemy’s Theorem).
In any cyclic quadrilateral the area of the rectangle contained by the two diagonals is equal to the sum of the area of the two rectangles contained by the two pairs of opposite sides.
Special Nomination:
Suppose ABCD is a cyclic quadrilateral whose two pairs of opposite sides are AB, CD and BC, AD. AC and BD are its diagonals. It is to be proved that, AC.BD = AB.CD + BC. AD.
Proof:
(Without loss of generality) we can assume that ZBAC is smaller than ∠DAC. Then we draw ∠DAP at point A with line segment AD, making it equal to ∠BAC so that AP intersects the diagonal BD at point P.
According to the construction, ∠BAC = ∠DAP.
Adding ∠CAP in both sides, we get
∠BAC + /CAP = LDAP + ZCAP Therefore, BAP = LCAD
Now between ΔABP and ΔACD
∠BAP = ∠CAD and ∠ABD = ∠ACD [angles on the same segment of the
circle]
and remaining ∠APB = remaining ∠ADC
ΔABP and ΔACD equiangular.
BP/CD = AB/ AC
So, AC .BP =AB. CD. = ……………(1)
Again, between ΔABC and ΔAPD
∠BAC = ∠PAD [by construction]
∠ADP =∠ACB [angles on the same segment of the circle]
and remaining ∠ABC = remaining ∠APD
ΔABC and ΔAPD equiangular.
AD/AC =PD/BC
Therefore, AC.PD = BC. AD ·………..(2)
Now adding equations (1) and (2), we get
AC BPAC PD AB CD + BC. AD =
Or, AC(BP + PD) = AB CD + BC. AD
But BP + PD = BD
So AC BD AB CD + BC. AD = [Proved]
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