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Symbol Defintion
Example : GT means greater than
Q01. Ortho-center
Defintion
- Three heights of triangle are concurrent at a poin O which is ortho-center
- Draw a triangle ABC
- Draw height AP perpendicualar to BC
- Draw height BQ perpendicualar to CA
- Draw height CR perpendicualar to AB
- Three height meets at one point O which is the orthocenter
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Q02. Pedal triangle : the feets of the heights form a triangle
- Draw a large triangle
- Draw AP perpendicular to BC
- Draw BQ perpendicular to CA
- Draw CR perpendicular to AB
- Hence three feets of the heights are P, Q, R
- Pedal triangle contains four similar triangle
- Triangle APQ is similar as ABC
- Angle CQP + angle AQP = 180 degrees (supplementary)
- Angle AQP + angle ABC = 180 degrees (concyclic)
- Hence angle CQP = angle ABC
- Angle C is in common for these two triangle
- Hence they are similar
- Similarly, triangle AQR is similar to triangle ABC
- Similarly, triangle AQR is similar to triangle ABC
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Q03. Points A, B, P, Q are concylic
Construction
- Draw a large triangle
- Draw AP perpendicular to BC
- Draw BQ perpendicular to CA
- Draw CR perpendicular to AB
- Hence three feets of the heights are P, Q, R
- Angle APB = 90 degrees
- Angle AQB = 90 degrees
- Hence points A, B, P, Q are concylic if AB as diameter to draw a circle
- Similarly, B, C, Q, R are concylic if BC as diameter
- Similarly, C, A, R, P are concylic if BC as diameter
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Q04. Angles of pedal triangle PQR and angles of triangle ABC
Angles of triangle PQR and angles of triangle ABC
- Angle PQR = 180 - 2*B
- Angle QRP = 180 - 2*C
- Angle RPQ = 180 - 2*A
- Since angle CQP = angle ABC and angle AQR = angle ABC
- Angle PQR = 180 - angle CQP - angle AQR
- Angle PQR = 180 - 2*angle ABC
- Similarly we prove ther two cases
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Q05. Sides of pedal triangle PQR
Relation with triangle ABC
- PQ = R*sin(2*C)
- QR = R*sin(2*B)
- RP = R*sin(2*A)
- Triangle PQA : circum-circle is circle with diameter = AB
- Hence PQ = AB*sin(PAQ)
- Angle PAQ + angle C = 90 degrees
- Hence PQ = AB*cos(C)
- Using sine law AB = c = 2*R*sin(C) where R is the circum-radius of triangle ABC
- Hence PQ = 2*cos(C)*sin(C) = R*sin(2*C)
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Q06. Area of triangle PQR
- PQ = R*sin(2*C)
- QR = R*sin(2*B)
- Angle PQR = 180 - 2*B)
- Area = (PQ*QR*sin(PQR))/2
- Area = (R*sin(2*A)*R*sin(2*b)*sin(2*b))/2
- Area = ((R^2)*sin(2*A)*sin(2*b)*sin(2*b))/2
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Q07. Circum-radius of triangle PQR
- In triangle PQR and by sine law
- RR = PQ/(2*sin(PQR))
- RR = (R*sin(2*C))/(2*sin(180 - 2*C)
- RR = (R*sin(2*C))/(2*sin(2*C)
- RR = R/2 where RR is the circum-radius of triangle PQR
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Q08. Orthocenter to sides of traingle ABC
Distance from O to sides
- OR = 2*R*cos(A)*cos(B)
- OP = 2*R*cos(B)*cos(C)
- OQ = 2*R*cos(C)*cos(A)
- In triangle AOR : OR = AR*tan(OAR)
- In triangle ARC : AR = AC*cos(A)
- In triangle ABP : AR = tan(OAR) = tan(90 - B) = cot(B)
- Hence OR = AC*cos(A)*cot(B)
- Hence OR = AC*cos(A)*cos(B)/sin(B)
- In triangle ABC, AC/sin(B) = 2*R
- Hence OR = 2*R*cos(A)*cos(B)
- Similarly, we prove other two
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Q09. Orthocenter to vertices of traingle ABC
Distance from O to sides
- OA = 2*R*cos(A)
- OB = 2*R*cos(B)
- OC = 2*R*cos(C)
- In triangle POC : OC = CP*sec(OCP)
- In triangle APC : CP = AC*cos(C)
- In triangle RBC : Angle OCP = 90 - B
- Hence OC = CP*csc(B)
- Hence OC = AC*cos(C)*csc(B)
- Hence OR = AC*cos(C)/sin(B)
- In triangle ABC, AC/sin(B) = 2*R (Sine law)
- Hence OR = 2*R*cos(C)
- Similarly, we prove other two
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Q10. Coordinates of Center
Let AB in horizton direction and A is the origin
- x = AR = b*cos(A)
- y = OR = 2*R*cos(A)*cos(B)
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Q11. References
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