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Congruency of Triangles

Two figures are said to be congruent if, when placed one over the other, they completely overlap each other. They would have the same shape, the same area and will be identical in all respects.
 
So, we can say that all congruent triangles are similar triangles, but vice-versa is not always true.
 
Rules for two triangles to be congruent
  1. S – S – S
     
    If in any two triangles, each side of one triangle is equal to a side of the other triangle, the two triangles are congruent. This rule is S – S – S rule.
  2. S – A – S
     
    In ∆ ABC and ∆ ABD,
     
    AB = AB (common side)
     
    ABC = ∠ BAD (given)
     
    BC = AD (given)
     
    Description: 11349.png
     
    Thus by rule S – A – S the two triangles are congruent.
     
    This rule holds true, when the angles that are equal have to be included between the two equal sides
     
    (i.e., the angle should be formed between the two sides that are equal).
  3. A – S – A
     
    In ∆ ABC and ∆ ADE,
     
    ∠ ACB =  AED (given)
     
    ∠ BAC = ∠ DAE (common angle)
     
    BC = DE (given)
     
    Thus by rule A – S – A the two triangles are congruent.
     
    For this rule, the side need not be the included side.
     
    Description: 6581.png 
     
    A – S – A can be written as A – A – S or S – A – A also.
  4. R – H – S
     
    This rule is applicable only for right-angled triangles.
     
    If two right-angled triangles have their hypotenuse and one of the sides as same, then the triangles will be congruent.

Similarity of the Triangles

If we take two maps of India of different sizes (breadths and lengths), then the map of all the 28 states of India will cover proportionally the same percentage area in both the maps.
 
Lets see this in geometry:
 
Criteria for similarity of two triangles
 
Description: P-332-1.tif
 
Two triangles are similar if (i) their corresponding angles are equal and/or (ii) their corresponding sides are in the same ratio. That is, if in two triangles, ABC and PQR, (i) A = P, B = Q, C = R, and/or (ii) Description: 6566.png the two triangles are similar.
 
All regular polygons of the same number of sides such as equilateral triangles or squares, are similar. In particular, all circles are also similar.

Theorems for Similarity

  1. If in two triangles, the corresponding angles are equal, then their corresponding sides will also be proportional (i.e., in the same ratio). Thus the two triangles are similar.
     
    This property is referred to as the AAA similarity criterion for two triangles.
     
    Corollary If two angles of a triangle are respec­tively equal to two angles of another triangle, then the two triangles are similar. This is referred to as the AA similarity criterion for the two triangles. It is true due to the fact that if two angles of one triangle are equal to the two angles of another tri­angle, then the third angle of both the triangles will automatically be the same.
  2. If the corresponding sides of two triangles are pro­portional (i.e., in the same ratio), their correspond­ing angles will also be equal and so the triangles are similar. This property is referred to as the SSS similarity criterion for the two triangles.
  3. If one angle of a triangle is equal to one angle of the other and the sides including these angles are proportional, then the triangles are similar. This property is referred to as the SAS similarity criterion of the two triangles.
  4. The ratio of the areas of the two similar triangles is equal to the ratio of the squares of their corre­sponding sides.
  5. If a perpendicular is drawn from the vertex of the right angle of a right angled triangle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each other.

Similar Polygons

Two polygons of the same number of sides are similar, if
  1. Their corresponding angles are equal (i.e., they are equiangular) and
  2. Their corresponding sides are in the same ratio (or proportional).
This can be seen in the figures given below:
Description: P-332-2.tif
 
Example-1
∆ABC is a right angled triangle BD ⊥ AC. If AD = 8 cm and DC = 2 cm, then BD = ?
Description: P-332-3.tif 
  1. 4 cm
  2. 4.5 cm
  3. 5 cm
  4. Cannot be determined
Solution
∆ ADB  ∆BDC
∴ Description: 6543.png
∴ BD2 = AD × DC = 8 × 2
∴ BD2 = 16
∴ BD = 4 cm
 
Important Result of this question BD2 = AD × DC can be used as a standard result also.
 
 
Example-2
Circles with radii 3, 4 and 5 units touch each other externally. If P is the point of intersection of the tangents to these circles at their point of contact, find the distance of P from the point of contacts of the circles.
Solution
Let A, B and C be the centres of the three circles. So, the point P will be the incentre of DABC and distance of P from the point of contacts of the circles will be the inradius (r).
 
So, Description: 6536.png
 
Sides of DABC will be 7 units, 8 units and 9 units.
 
So, Description: 6529.png
 

Some interesting facts

  1. If we draw regular polygons on all the sides of a right angled triangle, taking the sides of the triangle as one of the sides of the figures, then the area of the shaded portion is equal to twice the area of the figure on the hypotenuse. This will also hold true for semi-circles whose diameters will be the side of the right-angled triangle.
     
    Description: P-333-1.tif
     
    Area of the shaded region = 2 (area of figure on hypotenuse)
  2. In the figure below, two semi-circles are drawn with diameters equal to the sides of the right angled tri­angle. The area of the shaded region (the crescents) is equal to the area of the right-angled triangle. Area of the shaded region (is equal to) the area of the right angled triangle.
     
    Description: P-333-2.tif
  3. In the figure given below all triangles are equilateral triangles and circles are inscribed in these triangles. If the side of triangle ABC = a, then the side of triangle DEF Description: 6505.png and the side of triangle Description: 6498.png
     
    Description: P-333-3.tif
     
    In other words, we can say in order to obtain the side of the next inner triangle divide the side of the immediate outer triangle by 2. The same algorithm holds true for the inscribed circles.
  4. In the figure given below, if P is any point inside rectangle ABCD, then PA2 + PC2 = PB2 + PD2
     
    Description: P-333-4.tif

Some Important Points

  1. (a) In an acute angled triangle, the circumcentre lies inside the triangle.
     
    (b) In a right-angled triangle, the circumcentre lies on the middle point of the hypotenuse.
     
    (c) In an obtuse angled triangle, the circum­ centre lies outside the triangle.
  2. (a) In an acute angled triangle, the orthocenter lies inside the triangle.
     
    (b) In a right-angled triangle, the orthocenter lies on the vertex is where the right angle is formed i.e., the vertex opposite to the hy­ potenuse.
     
    (c) In an obtuse angled triangle, the orthocenter lies outside the triangle.
  3. In a right angled triangle the length of the median drawn to the hypotenuse is equal to half the hy­potenuse. This median is equal to the circumradius (R) of the right angled triangle.
  4. In the figure given below triangle ABC is a right angle triangle, right angled at B. Side AB mea­sures x units and BC measures y units. If a square (BDEF) the maximum area is inscribed in the triangle as shown, below then the side of the square is equal to Description: 6473.png
Description: P-333-5.tif
 
Example
In the figure given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB:CD = 3:1, then what is the ratio of CD: PQ?
Description: P-334-1.tif
Solution
Using the proportionality theorem, AB/PQ = BD/QD and PQ/CD = BQ/BD. Multiplying both these equations, we get AB/CD = BQ/QD = 3:1
 
Hence, CD/PQ = BD/BQ = 4 : 3.
 





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