A plane figure bounded by three lines in a plane is called a triangle. The figure below represents a ΔABC, with AB, AC andBC as the three line segments.

A triangle (denoted by the symbol △) is the simplest closed shape in geometry. It is a two-dimensional figure made by connecting three points that do not lie on the same straight line (non-collinear).

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.

An exterior angle is formed when a side of a triangle is extended beyond its vertex. The angle created between the extended side and the adjacent side of the triangle is the exterior angle.

Equilateral Triangle: A triangle in which all three sides are of equal lengths is called an equilateral triangle.

Isosceles Triangle: A triangle in which two sides are of equal lengths is called an isosceles triangle.    

Angle Sum Property of a Triangle:

Theorem: The sum of the angles of a triangle is 180°.

Construction: Draw a line XPY parallel to QR through the opposite vertex P.

Proof:

In △ PQR,
Sum of all angles of a triangle is 180°.
∠PQR + ∠PRQ + ∠QPR = 180°......(1)

Since XY is a straight line, it can be concluded that:

Therefore, ∠XPY + ∠QRP + ∠RPY = 180°.

But XPY || QR and PQ, PR are transversals.

So,
∠XPY = ∠PQR.....(Pairs of alternate angles)

∠RPY = ∠PRQ.....(Pairs of alternate angles)

Substituting ∠XPY and ∠RPY in (1), we get

∠PQR + ∠PRQ + ∠QPR = 180°

Thus, The sum of the angles of a triangle is 180°.

(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.