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Concept of Congruent and Similar

August 7, 2020
Category: Concept of Congruent and Similar

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☼ Concept of Congruent and Similar :–

■ Congruent Figures : –

Figures having the same shape, same measures and same areas are called congruent figures.

Above two circles are congruent. Above two hexagons are also congruent.

■ Similar Figures : –

Figures having the same shape but different measures and different areas are called similar figures.

Above two circles are similar. Above two hexagons are also similar. Also above two triangles.

■ Congruent Triangles :-

For a correspondence of two triangles, if the corresponding angles and measures of corresponding sides are congruent then this correspondence is called congruent.

In this case, those two triangles are called congruent triangle.

Here for $\Delta$ ABC & $\Delta$ PQR ,

m $\angle$ A = m $\angle$ P, m $\angle$ B = m $\angle$ Q, m $\angle$ C = m $\angle$ R

AB = PQ , BC = QR , AC = PR

$\large \Leftrightarrow$ $\Delta$ ABC $\cong$ PQR OR Correspondence ABC $\large \leftrightarrow$ PQR is congruent.

SAS Postulate :-

For a correspondence of two triangles, if two sides and the included angle of the one triangle is congruent to the corresponding two sides and included angle of the other triangle, then given two triangles are congruent.

e.g. Here,

AB = PQ , m $\angle$ A = m $\angle$ P, AC = PR $\large \Leftrightarrow$ $\Delta$ ABC $\cong$ $\Delta$ PQR

SAA Postulate :-

For the correspondence of two triangles, if two angles and a side of the one triangle are congruent to the corresponding two angles and a side of the other triangle, then given two triangles are congruent.

e.g. Here,

AB = PQ , m $\angle$ A = m $\angle$ P, m $\angle$ C = m $\angle$ R $\large \Leftrightarrow$ $\Delta$ ABC $\cong$ $\Delta$ PQR

ASA Postulate :-

For the correspondence of two triangles, if two angles and included side of the first triangle are congruent to the corresponding two angles and included side of the other triangle, then given two triangles are congruent.

e.g. Here,

m $\angle$ A = m $\angle$ P, AB = PQ , m $\angle$ B = m $\angle$ Q $\large \Leftrightarrow$ $\Delta$ ABC $\cong$ $\Delta$ PQR

SSS Postulate :-

For the correspondence of two triangles, if the measures of corresponding sides are congruent then this correspondence is called congruent.

e.g. Here,

AB = PQ , BC = QR , AC = PR $\large \Leftrightarrow$ $\Delta$ ABC $\cong$ $\Delta$ PQR

RHS Theorem :-

For a correspondence of two right angled triangle, if one side and hypotenuse is congruent to corresponding side and hypotenuse then that correspondence is called RHS and from these we gent two congruent triangles.

e.g. Here,

AB = PQ , m $\angle$ B = m $\angle$ Q = 90⁰, AC = PR $\large \Leftrightarrow$ $\Delta$ ABC $\cong$ $\Delta$ PQR

■ Similar Triangles :-

For a correspondence of two triangles, if the corresponding angles are congruent and the measures of the corresponding sides are in the same proportion, then this correspondence is called a similarity.

e.g. Here for $\Delta$ ABC & $\Delta$ PQR ,

m $\angle$ A = m $\angle$ P, m $\angle$ B = m $\angle$ Q, m $\angle$ C = m $\angle$ R

$\frac{AB}{PQ}= \frac{BC}{QR}=\frac{AC}{PR}$

$\large \Leftrightarrow$ $\Delta$ ABC ~ $\Delta$ PQR OR Correspondence ABC $\large \leftrightarrow$ PQR is similarity.

AA Postulate :-

If for any correspondence between two triangles two pairs of corresponding angles are congruent, then the correspondence is similarity.

e.g. Here for $\Delta$ ABC & $\Delta$ PQR ,

m $\angle$ A = m $\angle$ P, m $\angle$ B = m $\angle$ Q

$\large \Leftrightarrow$ $\Delta$ ABC ~ $\Delta$ PQR OR Correspondence ABC $\large \leftrightarrow$ PQR is similarity.

SAS Theorem :-

If for a correspondence between two triangles, two pairs of corresponding sides are proportional and the included angles are congruent, then the correspondence is a similarity.

e.g. Here for $\Delta$ ABC & $\Delta$ PQR ,

$\frac{AB}{PQ}=\frac{AC}{PR}$ , m $\angle$ A = m $\angle$ P

$\large \Leftrightarrow$ $\Delta$ ABC ~ $\Delta$ PQR OR Correspondence ABC $\large \leftrightarrow$ PQR is similarity.

SSS Theorem :-

For a given correspondence of two triangles, if the corresponding sides are in proportion, then the correspondence is a similarity.

e.g. Here for $\Delta$ ABC & $\Delta$ PQR ,

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

$\large \Leftrightarrow$ $\Delta$ ABC ~ $\Delta$ PQR

SSS Theorem :-

For a given correspondence of two triangles, if the corresponding sides are in proportion, then the correspondence is a similarity.

RHSI Theorem :-

For a given correspondence of two right angled triangle, if hypotenuse and one of the sides of a triangle is proportion to that of other triangle then the given correspondence is similar.

e.g. Here for $\Delta$ ABC & $\Delta$ PQR ,

$\frac{AB}{PQ}=\frac{AC}{PR}$ , m $\angle$ B = m $\angle$ Q = 90⁰

$\large \Leftrightarrow$ $\Delta$ ABC ~ $\Delta$ PQR

■ Formula Based on Similarity of Triangle :–

If $\Delta$ ABC ~ $\Delta$ PQR, then $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$ = $\frac{\Delta ABC}{\Delta PQR}$ = $\frac{AD}{PS}$ = $\frac{AE}{PT}$ = $\frac{AF}{PU}$ .

( Where, $\Delta$ ABC = Perimeter of $\Delta$ ABC. )

And $\frac{AB^{2}}{PQ^{2}}=\frac{BC^{2}}{QR^{2}}=\frac{AC^{2}}{PR^{2}}$ = $\frac{ ABC}{ PQR}$ = $\frac{AD^{2}}{PS^{2}}$ = $\frac{AE^{2}}{PT^{2}}$ = $\frac{AF^{2}}{PU^{2}}$ .

( Where, ABC = $\Delta$ = Area of $\Delta$ ABC. )

Also, Altitudes $\overline{AD}, \overline{PS}$

Angle Bisectors $\overline{AE}, \overline{PY}$

Medians $\overline{AF}, \overline{PZ}$

$\large \mapsto$ If $\overline{MN}$ is parallel to $\overline{BC}$ of $\Delta$ ABC, then

* $\frac{AM}{MB}=\frac{AN}{NC}$

* $\frac{AM}{AB}=\frac{AN}{NC}=\frac{MN}{BC}$

* $\frac{MB}{AB}=\frac{NC}{AC}$

$\large \mapsto$ In $\Delta$ ABC, G is the centroid and line l and through G is parallel to $\overline{BC}$ and intersect $\overline{AB}$ in M and $\overline{AC}$ in N, then

* $\frac{AM}{MB}=\frac{AN}{NC}=\frac{AG}{GD}=\frac{2}{1}$

* $\frac{AM}{MB}=\frac{AN}{NC}=\frac{MN}{BC}=\frac{AG}{AD}=\frac{2}{3}$

* $\frac{MB}{AB}=\frac{NC}{AC}=\frac{GD}{AD}=\frac{1}{3}$

$\large \mapsto$ In $\Delta$ ABC, line l is parallel to $\overline{BC}$ and intersect $\overline{AB}$ and $\overline{AC}$ in their respective mid–points, namely M and N, then

* $\frac{AM}{MB}=\frac{AN}{NC}=1$

* $\frac{AM}{AB}=\frac{AN}{NC}=\frac{MN}{BC}=\frac{1}{2}$

* $\frac{MB}{AB}=\frac{NC}{AC}= \frac{1}{2}$

$\large \mapsto$ In $\Delta$ ABC, medians $\overline{AD}$ and $\overline{BE}$ intersect at G such that $\overline{GK}$ ║ $\overline{DE}$ and K $\in$ $\overline{AC}$ , then

AC = 6 EK

$\large \mapsto$ In $\Delta$ ABC, D, E and F are respectively the mid–points of $\overline{AB},\overline{BC}$ and $\overline{AC}$ . Then

ABC = 4 $\times$ DEF

ADF = DBE = FBC = EFD

( Always Remember, ABC = $\Delta$ = Area of $\Delta$ ABC & $\Delta$ ABC = Perimeter of $\Delta$ ABC . )

$\large \mapsto$ If in $\Delta$ ABC, $\overline{BC}$ is produced to D, then m $\angle$ ACD = m $\angle$ A + m $\angle$ B.

In this case, m $\angle$ ACD is called the exterior angle of $\Delta$ ABC. And $\angle$ A and $\angle$ C are opposite interior angles of m $\angle$ ACD.

i.e. Exterior angle = Sum of its opposite interior angles

$\large \mapsto$ In $\Delta$ ABC, if $\overline{AM}$ is the bisector of $\angle$ BAC and $\overline{AN} \perp \overline{BC}$ , then

m $\angle$ MAN = $\frac{1}{2}$ ( m $\angle$ B – m $\angle$ C )

$\large \mapsto$ Pythagoras’ Law :

In a right angled ABC, if m $\angle$ B is a right angle, then

AB² + BC² = AC²

AC² = AB² + BC²

Some of the Pythagorean triplets are:

( 3, 4, 5 ); ( 5, 12, 13 ); ( 7, 24, 25 ); (8, 15, 17 ); ( 9, 40, 41 ); ( 11, 60, 61 );

( 12, 35, 37 ); ( 16, 63, 65 ), ( 20, 21, 29 )

$\large \mapsto$ In an acute angled $\Delta$ ABC, if m $\angle$ B is an acute angle, then

AC² < AB² + BC²

$\large \mapsto$ In an obtuse angled $\Delta$ ABC, if m $\angle$ B is an obtuse angle, then

AC² > AB² + BC²

$\large \mapsto$ In an acute angled $\Delta$ ABC, if m $\angle$ B is an acute angle, then

AC² = AB² + BC² – 2BC • BM

$\large \mapsto$ In an obtuse angled ABC, if mB is an obtuse angle, then

AC² = AB² + BC² + 2BC • BM

$\large \mapsto$ In a right angled ABC, if mB is a right angle, then

* BM² = AM $\times$ CM

* AB² = AM $\times$ AC

* BC² = CM $\times$ AC

$\large \mapsto$ If line l passes through the mid–points M and N of $\overline{AB}$ and $\overline{AC}$ respectively, then l ║ $\overline{BC}$ & MN = $\frac{1}{2}$ BC

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Concept of Congruent and Similar

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