Revision: Geometry Mathematics SSLC (English Medium) Class 8 Tamil Nadu Board of Secondary Education

Two figures are similar if they have the same shape but may differ in size.

• Same shape means: Corresponding angles are equal

• May differ in size means: Corresponding sides are proportional

Two figures are congruent if they have the same shape and size.

Important relation:

• Congruent figures are always similar
• Similar figures are not necessarily congruent

SAS Congruence criterion: If under a correspondence, two sides and the angle included between them of a triangle are equal to two corresponding sides and the angle included between them of another triangle, then the triangles are congruent.

ASA Congruence criterion: If under a correspondence, two angles and the included side of a triangle are equal to two corresponding angles and the included side of another triangle, then the triangles are congruent.

RHS Congruence criterion: If under a correspondence, the hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right-angled triangle, then the triangles are congruent.

Two triangles are similar if

• Their corresponding angles are equal, and
• Their corresponding sides are proportional.
• Symbolically:
  ΔABC ∼ ΔPQR (read as “ABC is similar to PQR”).

In similar triangles, the angles opposite to proportional sides are the corresponding angles, and so, they are equal. 

• ∠A = ∠P

• ∠B = ∠Q

• ∠C = ∠R

In similar triangles, the sides opposite to equal angles are said to be the 
corresponding sides. 

ΔABC ∼ ΔPQR

\[\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}\]

The line segment joining a vertex of a triangle to the midpoint of its opposite side is called a median of the triangle.

• The point where all three medians meet is called the centroid.

An altitude of a triangle is a straight line from one vertex, drawn perpendicular (at 90°) to the opposite side (or to its extension)

• Orthocentre: The altitudes of a triangle pass through exactly one point; that means they are concurrent. The point of concurrence is called the orthocentre.

A circumcircle is a circle that passes through all three vertices of a triangle. The three vertices lie on the boundary of the circle.

The circumcenter is the center point of the circumcircle. It is the unique point where all three perpendicular bisectors of the triangle's sides meet.

• The circumcenter is equidistant from all three vertices of the triangle.

The circumradius is the radius of the circumcircle. It is the distance from the circumcenter to any vertex of the triangle.

Given: A(ΔPQR) in which QX is the bisector of ∠Q and RX is the bisector of ∠R.

XS ⊥ QR and XT ⊥  PQ.

We need to prove that:

1. ΔXTQ ≅ ΔXSQ.
2. PX bisects angle P.

Construction: Draw XZ ⊥ PR and join PX.

i. In ΔXTQ and ΔXSQ,

∠QTX = ∠QSX = 90°  ...[XS ⊥ QR and XT ⊥  PQ]

∠TQX = ∠SQX    ...[QX is bisector of ∠Q]

QX = QX    ...[Common]

∴ By Angle-Side-Angle Criterion of congruence,

ΔXTQ ≅ ΔXSQ

ii. The corresponding parts of the congruent triangles are congruent.

∴ XT = XS   ...[c.p.c.t.]

In ΔXSR and ΔXRZ

∠XSR = ∠XZR = 90°   ...[XS ⊥ QR and ∠XSR = 90°]

∠XRS = ∠ZRX      ...[RX is bisector of ∠R]

RX = RX    ....[Common]

∴ By Angle-Angle-Side criterion of congruence,

ΔXSR ≅ ΔXRZ

The corresponding parts of the congruent triangles are congruent.

∴ XS = XT    ...[c.p.c.t.] 

From (1) and (2)

XT = XZ                    

In ΔXTP and ΔPZX

∠XTP = ∠XZP = 90°    ....[Given]

XP = XP         ....[Common]

XT = XZ               

∴ By Right angle-Hypotenuse-side criterion of congruence,

ΔXTP ≅ ΔPZX

The corresponding parts of the congruent triangles are
congruent.

∴ ∠TPX = ∠ZPX    ...[c.p.c.t.]

∴ PX bisects ∠P.

Theorem (ASA congruence rule) :  Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle. 
Proof : We are given two triangles ABC and DEF in which: 
∠ B = ∠ E, ∠ C = ∠ F 
and BC = EF 
We need to prove that   ∆ ABC ≅ ∆ DEF
For proving the congruence of the two triangles see that three cases arise.

Case (i) : Let AB = DE in following fig.

You may observe that
AB = DE (Assumed) 
∠ B = ∠ E (Given)
BC = EF (Given) 
So, ∆ ABC ≅ ∆ DEF   (By SAS rule)

Case (ii) : 
Let if possible AB > DE. So, we can take a point P on AB such that PB = DE. Now consider ∆ PBC and ∆ DEF  in following fig. 

Observe that in ∆ PBC and ∆ DEF,
PB = DE (By construction) 
∠ B = ∠ E (Given)
BC = EF (Given)
So, we can conclude that: 
∆ PBC ≅ ∆ DEF, by the SAS axiom for congruence.
Since the triangles are congruent, their corresponding parts will be equal. 
So, ∠ PCB = ∠ DFE
But, we are given that
∠ ACB = ∠ DFE So, ∠ ACB = ∠ PCB
This is possible only if P coincides with A.
or, BA = ED 
So, ∆ ABC ≅ ∆ DEF   (by SAS axiom)

Case (iii) :  If AB < DE, we can choose a point M on DE such that ME = AB and repeating the arguments as given in Case (ii),  we can conclude that AB = DE and so, ∆ ABC ≅ ∆ DEF. 
You know that the sum of the three angles of a triangle is 180°. So if two pairs of angles are equal, the third pair is also equal (180° – sum of equal angles).
So, two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal. We may call it as the AAS Congruence Rule.

(i) Applying Pythagoras theorem in ΔAMD, we obtain

AM² + MD² = AD² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC² [Using equation (1)]

Using the result, DC = `"BC"/2`, we obtain

`"AD"^2+(("BC")/2)^2 + 2"MD".(("BC")/2) = "AC"^2`

`"AD"^2+(("BC")/2)^2 + "MC" xx "BC" = "AC"^2`

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB² = AM² + MB²

= (AD² − DM²) + MB²

= (AD² − DM²) + (BD − MD)²

= AD² − DM² + BD² + MD² − 2BD × MD

= AD² + BD² − 2BD × MD

= `"AD"^2+(("BC")/2)^2 - 2(("BC")/2) xx "MD"`

= `"AD"^2 + ("BC"/2)^2 - "BC" xx "MD"`

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AM² + MB² = AB² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC² … (2)

Adding equations (1) and (2), we obtain

2AM² + MB² + MC² = AB² + AC²

2AM² + (BD − DM)² + (MD + DC)² = AB² + AC²

2AM²+BD² + DM² − 2BD.DM + MD² + DC² + 2MD.DC = AB² + AC²

2AM² + 2MD² + BD² + DC² + 2MD (− BD + DC)

= AB² + AC²

= `2("AM"^2 + "MD"^2) + (("BC")/2)^2 + (("BC")/2)^2 + 2"MD" ((-"BC")/2 + ("BC")/2) = "AB"^2 + "AC"^2`

`"2AD"^2 + ("BC"^2)/2 = "AB"^2 + "AC"^2`

Applying Pythagoras theorem in ΔAMD, we obtain

AM² + MD² = AD² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC² [Using equation (1)]

Using the result, DC = `"BC"/2`, we obtain

`"AD"^2+(("BC")/2)^2 + 2"MD".(("BC")/2) = "AC"^2`

`"AD"^2+(("BC")/2)^2 + "MC" xx "BC" = "AC"^2`

Draw perpendicular BD from the vertex B to the side AC. A – D

In right-angled ΔABC

seg BD ⊥ hypotenuse AC.

∴ By the similarity in right-angled triangles

ΔABC ~ ΔADB ~ ΔBDC

Now, ΔABC ~ ΔADB

∴ `"AB"/"AD" = "AC"/"AB"`   ...(c.s.s.t)

∴ AB² = AC × AD   ...(1)

Also, ΔABC ~ ΔBDC

∴ `"BC"/"DC" = "AC"/"BC"`   ...(c.s.s.t)

∴ BC² = AC × DC   ...(2)

From (1) and (2),

AB² + BC² = AC × AD + AC × DC

= AC × (AD + DC)

= AC × AC   ...(A – D – C)

∴ AB² + BC² = AC²

i.e., AC² = AB² + BC²

Given: In ΔPQR, ∠PQR = 90°.

To prove: PR² = PQ² + QR².

Construction:

Draw seg QS ⊥ side PR such that P–S–R.

Proof: In ΔPQR,

∠PQR = 90°   ...(Given)

Seg QS ⊥ hypotenuse PR   ...(Construction)

∴ ΔPSQ ∼ ΔQSR ∼ ΔPQR   ...(Similarity of right-angled triangles)   ...(1)

ΔPSQ ∼ ΔPQR   ...[From (1)]

∴ `"PS"/"PQ" = "PQ"/"PR"`   ...(Corresponding sides of similar triangles are in proportion)

∴ PQ² = PS × PR   ...(2)

ΔQSR ∼ ΔPQR   ...[From (1)]

∴ `"SR"/"QR" = "QR"/"PR"`   ...(Corresponding sides of similar triangles are in proportion)

∴ QR² = SR × PR   ...(3)

Adding (2) and (3), we get

PQ² + QR² = PS × PR + SR × PR

∴ PQ² + QR² = PR(PS + SR)

∴ PQ² + QR² = PR × PR   ...(P–S–R)

∴ PQ² + QR² = PR² OR PR² = PQ² + QR².

Given: In triangle PQR, ∠PQR = 90° and S is the mid-point of QR.

To prove: QR² = 4(PS² – PQ²)

in right-angled ΔPQS, by Pythagoras theorem,

PQ² + QS² = PS²

⇒ QS² = PS² – PQ²   .......(i)

Since S is the mid-point of side QR,

∴ QS = `(QR)/2`

Substituting the value of QS in equation (i),

`((QR)/2)^2 = PS^2 - PQ^2`

`(QR^2)/4 = PS^2 - PQ^2`

QR² = 4(PS² – PQ)²

Hence proved.

• AA / AAA → two angles equal

• SAS → included angle equal + sides proportional

• SSS → all sides proportional