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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
- Draw a large triangle ABC
- Draw AD perpendicular to BC
- Draw BE perpendicular to CA
- Draw CF perpendicular to AB
- Draw line parallel to AB and pass C
- Draw line parallel to BC and pass A
- Draw line parallel to CA and pass B
- Three parallel lines meet at P, Q, R
- P, C, Q are colinear
- Q, A, R are colinear
- R, B, P are colinear
- Hence PQR is a triangle
- CQ parallel to AB (construction)
- AQ parallel to BC (construction)
- Hence ABCQ is a parallelogram
- Hence QC = AB
- Similarly ABPC is a parallelogram
- Hence PC = AB
- Since CF perpendicular to AB and PQ parallel to AB
- Hence PC = QC and CF perpendiculer to PQ
- Hence CF bisects PQ
- Same as above
- Same as above
- Since CF, AD and BE are are bisectors of PQR
- Hence CF, AD and BE are concurrent at O (circum-center)
- But AD, BE nad CF are heigths of triangle ABC
- Hence heights of triangle are concurrent
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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
- Draw a large triangle
- Draw orthod-center O
- Draw circum-center D
- Draw COF perpendicular to AB
- Draw DP Perpendicular to AB as bisector
- Join mid point M of CO and mid point N of AO
- In triangle AOC we have MN = AC/2 and MN parallel to AC
- Join mid point R of BC and mid point P of AB
- In triangle ABC we have PR = AC/2 and PR parallel to AC
- Hence we can prove that triangle MON is similar to QOR
- PR parallel to MN
- DP parallel to MO
- NO parallel to DR
- Hence DP = MO = CO/2
- Hence CO : DP = 2 : 1
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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.
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Q09.
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Q10. References
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