Tuesday, 23 December 2025
In a triangle, in which ratio orthocenter divides the altitude?
AO/OP = cos A / (cos B cos C)
Using
1) A = 180 degree - (B + C)
2) cos [180 -(B + C)] = -cos(B + C) and cos (B + C) = cosBCosC-sinBsinC
In triangle ABC if AP is perpendicular to BC such that B-P-C i.e. P is the foot of perpendicular from vertex A on side BC and O is orthocenter then AO/OP = tanB tan C - 1
From this we understand that in a triangle the ratio in which the orthocenter divides the altitude does not depend on the lengths of the sides but depends on the angles of the triangle.
The vertex from which altitude is drawn i.e. the angle A
1) will be acute if the ratio is positive
2) will be 90 degrees if the ratio is zero
3) will be an obtuse angle if the ratio is negative.
Conversely if the ratio is
1) positive the angle A will be an acute angle
2) zero angle A will be 90 degrees
3) negative the angle A will be an obtuse angle.
Activity to verify Angle bisector property or theorem
Activity:
In a triangle ABC, AB= 5 CM, BC = 8 CM, AC = 7 CM.
Angle B = 60 degrees. Bisect angle B. Let ray BD bisect angle B such that A-D-C
I suggest to use this triangle for verifying the angle bisector property/theorem.
Advantage: 60 degrees angle can be easily bisected , even with protractor.
Approximately AD = 2.7 CM ; DC = 4.3 CM
Students should know tables upto 30, so they can divide by 27.
Make it AD/DC= 4.3/2.7=43/27
Another thing I teach them is approximation.
43 can be approaximated to 43.2 Since 27 is divisible by 3, and since sum of three consecutive number is always divisible by 3.
Therefore we put 43.2. Now divide numerator and denominator by 3. AD/DC=14.4/9=1.6
Some sharp students may ask why not 43.5 since 5,4,3 are also consecutive three numbers? then you have to reply 43.5 = 44
where as 43.2 will be equal to 43 as per approximation rule.
Find area of triangle in a rectangle, its sides; given
In rectangle ABCD, points E and F are on sides AD and DC such that A-E-D and D-F-C. Areas of triangles are as follows:
Tri. ABE = 180 sq. cm. ;
Tri. BCF = 384 sq. cm. ;
Tri. DEF = 60 sq. cm.;
AB, BC, AE, FC, ED, DF are all positive integers. AB > BC
Find:
1) Area of [] rectangle ABCD
2) Area of Tri. BEF.
3) side AB given AB - BC = 16 cm
4) side BC
5) Also find sides of the triangle BEF
6) You may verify area of the tri. obtained in 2) by Heron's formula. Do you find the lengths of segments (dimensions of triangles and reectangle) wonderful?
It is the grace of Saint Saibaba, Shirdi, Maharashtra, India and Goddess Mahalaksmi that inspired me this sum.
Do you think an orthocentre of an equilateral triangle only will divide its altitude in the ratio 2:1, then you are wrong! Please read this post.
A = ( x1 , y1) , B(0, 0) & C(x3, 0)
O is the orthocenter and it divides perpendicular to BC which is AP in the ratio 2:1.
That is AO / OP = 2 / 1
That is coordinates of O (x1, y1 /3)
Slope of AC= (y1-0)/(x1 - X3) = y1 /(x1-x3)
Slope of BH = (y1/3 -0) /(x1-0) = y1/(3*x1)
AC perpendicular to BH,
[y1/(x1-x3) ]*[(y1/(3x1) ]= -1
(y1) ^2=-3x1(x1-x3)
=3x1(x3-x1)
Take x1=3, x3=4
so y1= + / - 3
AB= Sqrt(10), AC = 3* Sqrt(2) = Sqrt(18) , BC = 4
AP = 3
AO = 2, OP = 1
Here we are discussing abpout only one altitude to get internally divided in the ratio 2:1.
Thank you.
This post I am able to write due to grace of Saibaba and SHree Gajanan Mahraj, Shegaon, Maharashtra.
Thursday, 5 October 2023
My Sum on Cevian Theorem
"In Δ ABC, B-P-C, BP = 4, PC = 8, A-Q-C, AQ = QC = 6. BQ intersects AP at point X. If CX is drawn and extended to intersect AB at point R , find length of PR.
"
Wednesday, 4 October 2023
Triple Heronian Triangle with all the three integer sides and all the three corresponding integer heights (altitudes)
Friday, 15 September 2023
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