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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 AD perpendicualar to BC
- Draw height BE perpendicualar to CA
- Draw height CF perpendicualar to AB
- Three height meets at one point O which is the orthocenter
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Q02. Prove that three heights meet at one point
Diagram
- Draw a large triangle ABC
- Draw AP perpendicular to BC
- Draw BQ perpendicular to CA
- Draw CR perpendicular to AB
- Hence three heights are concurrent at O
- Use equations of two lines to find intersection
- Assume coordinates of A, B, C are given
- Then slope of AB can be found
- Hence slope of CR = -1/(slope of AB). Equation of CR is obtained
- Then slope of BC can be found
- Hence slope of AP = -1/(slope of BC). Equation of AP is obtained
- Solve equations AP and CR to get point of intesection O
- Then slope of CA can be found
- Hence slope of BQ = -1/(slope of CA). Equation of BQ is obtained
- Solve equations AP and BQ to get point of intesection O'
- Point O and O' shpuld have same coordinates
- See GE 03 03 : Ortho-center theory
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Q03. Property 1 : 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. Pedal triangle : the feets of the heights form a triangle
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
- 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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Q05. Ortho-center O of traingle ABC is also incenter of triangle PQR
Prove point O is in-center of triangle PQR
- BQ is bisector of angle PQR
- Triangle CQP is similar triangle AQR
- Hence angle AQR = angle CQP
- Angle AQR + RQB = 90
- Angle BQP + PQC = 90
- Hence BQR = BQP
- Hence BQ is bisector of angle PQR
- Similarly, AP is besector of angle QPR
- Similarly, CR is besector of angle PRQ
- Hence O is in-center of PQR
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Q06. Ortho-center : relation with circum-center
Construction
- Draw a large triangle
- Draw orthod-center O
- Draw circum-center U
- Draw COF perpendicular to AB
- Draw UR Perpendicular to AB
- Construction method : By measurement
- Geometric method
- Join mid point M of AO and mid point N of BO
- In triangle BOC we have MN = BC/2 and MN parallel to BC
- Join mid point R of AB and mid point Q of CA
- In triangle ABC we have QR = BC/2 and QR parallel to BC
- Hence we can prove that triangle MON is similar to QOR
- Hence ER = ON = AO/2
- Hence AO : UR = 2 : 1
- Use distance formula find AO if coordinates A and O are known
- Use distance formula find UR if coordinates U and R are given
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Q07. 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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Q08. 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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Q09. Question : If coordiantes of A, B, C are given
- How to find coordinare of orhto-center O ?
- Find intersection of CR and BQ
- The slope of CR = -1/(slope of AB)
- The slope of BQ = -1/(slope of CA)
- Hence equations of BQ and CR can be found
- Solve two equations we can find O
- How to find coordinate of R ?
- R is mid-point of point A and B
- How to find coordinate of circum-center U ?
- Find intersection of UQ and UR and Q is mid point of C and A
- The slopes of UQ and UR are known.
- Hence equations of UQ and UR can be found
- Solve two equations we can find U
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Q10. References
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