Mathematics Academy

Showing posts with label Triangle. Show all posts

Saturday, September 14, 2019

त्रिभुज का बहिष्कोण (Exterior Angle of a Triangle)

त्रिभुज का बहिष्कोण

त्रिभुज ABC की भुजा BC को D तक बढ़ाने पर त्रिभुज का बहिष्कोण ∠ACD प्राप्त होता है।

चित्र से स्पष्ट है कि,∠3 और ∠4 रैखिक युग्म बना रहे हैं,अत: हम लिख सकते हैं :
∠3+∠4=180° …..(1)
हम जानते हैं कि त्रिभुज के तीनों अंत:कोणों का योग 180° होता है , अत:
∠1+∠2+∠3= 180° …..(2)
समीकरण (1) व (2) से
∠1+∠2+∠3= ∠3+∠4
या ∠1+∠2= ∠4

परिणाम : यदि किसी त्रिभुज की एक भुजा बढ़ाई जाए , तो इस प्रकार बना बहिष्कोण दोनों अंत: विपरीत कोणों के योग के बराबर होता है। यहां यह भी स्पष्ट होता है कि किसी त्रिभुज का बहिष्कोण अपने दोनों अंत: विपरीत कोणों में से प्रत्येक से बड़ा होता है।

Exterior Angle of a Triangle

By extending the side BC of a triangle ABC , we get exterior angle ∠ACD of the triangle.
It is clear from the figure that ∠3 and ∠4 form a linear pair , so we can write that :
∠3+∠4=180° …..(1)
We know that , the sum of interior angles of a triangle is 180° , so
∠1+∠2+∠3= 180° …..(2)
From equations (1) and (2)
∠1+∠2+∠3= ∠3+∠4
Or ∠1+∠2= ∠4

Result: If we extend one side of a triangle, then the exterior angle so formed is equal to the sum of two interior opposite angles. It is clear from the above that the exterior angle of a triangle is greater than the corresponding two interior opposite angles.

Sunday, December 13, 2015

Nine Point Conics

Nine point circle of a triangle ABC is a circle passing through the mid points of sides of the triangle (G, H, I), feet of the perpendicular (D, E , F) drawn from the vertex to the opposite sides and the mid-points (K , L, M) of the distance from the orthocenter (N) to the three vertices of the triangle.

The concept of a nine point circle can be generalized to a nine point ellipse or a nine point hyperbola if we consider a general cevian instead of altitude. A cevian is any segment drawn from the vertex of a triangle to the opposite side. Cevians with special properties include altitudes, angle bisectors, and medians.

Consider three concurrent cevians with cevian point P, locate mid-points E, F and G of the segments from cevian point to the vertices of the triangle. Also locate the feet of the cevian K , N and L. If we draw a conic through any five of the above points, we will get an ellipse and it will also pass through the sixth point.

Now locate the mid points of the three sides of the triangle, we will find that these points also fall on the ellipse constructed above. The conic remains an ellipse when the feet of cevians lie on the sides of the triangle and converts to a nine point hyperbola when the feet of the cevians lie on the extensions of the sides.

Now locate centroid (G₁) of the triangle ABC and centre (N₁) of the conic , interestingly , the cevian point P , G₁ and N₁ lie on the same straight line with N₁P = 3 N₁G₁ . This is also the generalization of Euler Line.

Friday, November 9, 2012

GeoGebra Tutorial - 6 Triangle and Angles

In this tutorial we will construct a triangle and measure its internal angles.

Click on File Menu and select New. (This will open a new GeoGebra window)
Click on the drop down arrow of the Line through Two Points and choose the Segment between Two Points button

Construct a triangle using Segment between Two Points button. You will see in the Algebra Window , the three segments (sides of triangle) are represented by their lengths.

To see which length corresponds to which side , click on the Move Tool (arrow) and hover the mouse over one of the sides to see the corresponding length light up.
Now to measure internal angles of the triangle , select the Angle Button and click once on vertices B , A and C in that order. This will give you measure of ∠BAC. (Make sure that you select the vertices in clockwise direction. Anti-clockwise selection will give you reflex angle.

Similarly, measure ∠CBA and ∠ACB.
Now, drag the vertices using Move Button (Arrow). If you drag enough, you will notice reflex angles appearing. To stop this, right click on an angle in Algebra or Geometry Window, and choose Object Properties.Under Basic Tab un-check the Allow Reflex Angle check box and close the Object Properties box.

Check your settings have been applied by dragging the vertices of your triangle.

Sunday, August 12, 2012

Practice Qustions - Angle Sum Property

Friday, July 27, 2012

Exterior Angle Sum Property of a Triangle

A triangle has three corners, called vertices. The sides of a triangle that come together at a vertex form an angle. This angle is called the interior angle. In the figure below, the angles a,b and c are the three interior angles of the triangle. We know that the sum of interior angles of a triangle is 180°. You may visit the links http://mathematicsbhilai.blogspot.in/2012/01/triangle-angle-sum-property-ii.html and http://mathematicsbhilai.blogspot.in/2011/05/triangle-angle-sum.html

An exterior angle is formed by extending one of the sides of the triangle; the angle between the extended side and the other side is the exterior angle. In the figure, angle d is an exterior angle.

In the above figure ∠ACD is an exterior angle of Δ ABC.

Because ∠a +∠b +∠c = 180°, and ∠b +∠d = 180°, we can see that that ∠d =∠a +∠c. This is stated as a theorem. An exterior angle of a triangle is equal to the sum of the two opposite (nonadjacent) interior angles. This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Thursday, May 31, 2012

Triangle and Incentre

Let ABC be a triangle in which AB = AC and let I be its in centre. Suppose BC = AB + AI. Find ∠BAC.

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Extend CA to D such that AD = AI , then CD = CB by the hypothesis.
Hence ∠CDB = ∠CBD = 90° - (C/2) using angle sum property in triangle BCD.

Using angle sum property in triangles ABC and ABI we have ∠AIB = 90° + (C/2).
Thus ∠AIB + ∠ADB = 90° - (C/2) + 90° + (C/2) = 180°.

Hence ADBI is a cyclic quadrilateral. This implies that ∠ADI = ∠ABI = (B/2)

Now,triangle ADI is isosceles,as AD = AI , this gives ∠DAI = 180°-2(∠ADI)=180° - B.
Thus ∠CAI = B which gives A = 2B. As ∠C = ∠B , we get 4B = 180° and hence B = 45°.
Thus we get A = 2B = 90°

Wednesday, May 30, 2012

Solution of Triangle using Circumcircle

In triangle ABC , let D be the mid point of BC. If ∠ADB = 45° and ∠ACD = 30° , determine ∠BAD.
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Draw BL perpendicular to AC and join L to D. Since ∠BCL = 30° , we get ∠CBL = 60°.
Since BLC is a right triangle with ∠BCL = 30° , we have BL = BC /2 = BD. Thus in triangle BLD , we observe that BL = BD and ∠DBL= 60°.

This implies that BLD is an equilateral triangle hence LB = LD. Using ∠LDB = 60° and ∠ADB=45° , we get ∠ADL = 15°. But ∠DAL = 15° , thus LD = LA. Hence we have LD = LA = LB.

This implies that L is the circumcenter of the triangle BDA , thus ∠BAD = ∠BLD / 2 = 60° / 2 = 30°

Saturday, May 26, 2012

Circumcircles through a common point

Let ABC be a triangle and let P, Q, R be any points on the sides BC, CA, AB, respectively. Then the circumcircles of ARQ, BP R, CQP pass through a common point. This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Sunday, May 6, 2012