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.

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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°

Solution of Triangle using Circumcircle

In triangle ABC , let D be the mid point of BC. If ∠ADB = 45° and ∠ACD = 30° , determine ∠BAD.
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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°

Poles and Strings

A pair of telephone poles d metres apart is supported by two cables which run from the top of each pole to the bottom of the other. The poles are 4 m and 6 m tall. Determine the height above the ground of the point, T , where the two cables intersect. What happens to this height as d increases? 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
Let the height of the poles QP = a and RS = b.
Let the distance PO = c. Let h be the height above ground of the point T.

Since ∆QPR and ∆TOR are similar , so a : d = h : (d – c) , or (d – c ) = dh / a.
Similarly , ∆SRP and ∆TOP are similar , so b : d = h : c . So , c = dh / b.

Eliminating c from the above two equations we get d = dh (1/a + 1/b) , which gives h = ab / (a+b)
If a = 4 m and b = 6 m then we get h = 12 / 5 m. Thus height of point T is 12/5 m , independent of d.This can be seen by dragging points P or Q and see the path taken by point T.

Right Triangle and Square on Hypetenuse

Let ABC be a right triangle with right angle at B. Let AC DE be a square drawn exterior to triangle ABC. If M is the center of this square, find the measure of ∠ MBC.
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Note that triangle MCA is a right isosceles triangle with ∠AMC = 90° and ∠MAC = 45°. Since ∠ABC = 90°, there is a circle k with diameter AC which also passes through points B and C. Chord CM of circle k subtend angles MAC and MBC on the same segment . Hence ∠MBC = ∠MAC = 45°

Medians and Isosceles Triangle

If two medians in a triangle are equal in length,then the triangle is isosceles.

Let medians AM = BN in ∆ ABC. Extend each median to AP and BQ so that M and N are the midpoints of AP and BQ, respectively. Hence , AM = MP and BN = NQ. By the property of bisecting diagonals, ABPC and ABCQ are parallelograms. Hence CP and CQ are each parallel and equal to AB. We conclude that C lies on QP and C is the midpoint of QP. 

Now , AM = BN , so  2AM = 2BN  , hence AP = BQ. This shows that ABPQ is a trapezium with equal diagonals. 

It is easy to see that such a trapezium is isosceles. One way to see this is to draw a line through A parallel to diagonal BQ, until it intersects line QP in point L. Thus, ABQL is a parallelogram, so ∠ALQ = ∠ABQ. On the other hand, ∆APL is isosceles since AL = BQ = AP; hence, ∠ ALQ = ∠ APQ. 

Finally, AB || QP implies ∠  APQ = ∠ BAP. We conclude that ∠ BAP = ∠ ABQ, and ∆ ABQ and ∆ BAP are congruent by two equal sides and angles between these sides (SAS Criteria) . Therefore, BP = AQ and our trapezium is isosceles. 

 Hence ∠AQC=∠ BPC. Finally , ∆ACQ and ∆ BCP are congruent by AQ = BP, CQ = CP and ∠ AQC = ∠ BPC (SAS Criteria).  We conclude that AC = BC and our original ABC is isosceles.

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

tan 75° Geometrically

To find the value of tan 75° geometrically. 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

Tangent Circles - 2

Given a circle with radius R. The circle is tangent to a line ‘a’ at point M. A number r is given. Construct a circle with radius r tangent to both the given circle and the given line ‘a’
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Tangent Circles - I

Given a circle with radius r. The circle is tangent to a line ‘a’ at point M. There is a point N on the circle. Construct a circle which is tangent to given circle at N and the line ‘a’
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Pappu's Theorem

If A1, B1, and C1 are three points on a line, and A2 , B2 and C2 are three points on a second line.If A1B2 meets A2B1 at X, and A1C2 meets A2C1 at Y, and B1C2 meets B2C1 at Z, then X, Y, and Z are collinear (on a straight line).

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Tetrahedron's Dual

A tetrahedron is a Platonic Solid composed of four triangular faces, three of which meet at each vertex. It has six edges and four vertices. A tetrahedron is also known as a "triangular pyramid".
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Net of Tetrahedron

Pythagoras Theorem and Semicircles

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Frequency Polygon

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Morley's Triangle

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Mean , Mode , Median

Histogram

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Visualising AM-GM-HM

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