Geometry Concepts

1. Triangle Angle & Side Properties

• Formula:

$$\boxed{\angle A+\angle B+\angle C=180^\circ}$$ $$\boxed{\text{Exterior angle}=\text{sum of two opposite interior angles}}$$

For an isosceles triangle, equal sides have equal opposite angles. Use when: Finding missing angles, classifying triangles, or applying basic triangle properties.

• Example: In (\triangle ABC), (\angle A=50^\circ) and (\angle B=60^\circ). Find (\angle C).

Solution:

$$\angle C=180^\circ-(50^\circ+60^\circ)$$ $$\boxed{\angle C=70^\circ}$$

All three angles are less than (90^\circ), so the triangle is acute.

       A
      / \
 50° /   \ 70°
    /     \
   /_______\
  B   60°  C

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2. Exterior Angle of a Triangle

• Formula:

$$\boxed{\text{Exterior angle}=\text{sum of two opposite interior angles}}$$

Also,

$$\boxed{\text{Exterior angle}=180^\circ-\text{adjacent interior angle}}$$

Use when: A triangle has an extended side and an exterior angle is given.

• Example: Two opposite interior angles of a triangle are (45^\circ) and (65^\circ). Find the exterior angle.

Solution:

$$E=45^\circ+65^\circ$$ $$\boxed{E=110^\circ}$$

       A
      / \
 45° /   \ 65°
    /     \
   B-------C──────
              110°

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3. Similar Triangles

• Formula: If

$$\triangle ABC\sim\triangle DEF,$$

then

$$\boxed{\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}=k}$$

and

$$\boxed{\frac{\text{Area}_1}{\text{Area}_2}=k^2}.$$

Use when: Two triangles have equal corresponding angles or proportional corresponding sides.

• Example: A (3,4,5) triangle is similar to another triangle whose longest side is (15). Find the larger triangle’s area.

Solution: Original longest side (=5).

Scale factor:

$$k=\frac{15}{5}=3$$

Original area:

$$A=\frac12(3)(4)=6$$

Larger area:

$$6(3^2)=54$$ $$\boxed{54\text{ square units}}$$

Small:       Large:
  /|           /|
 / |  3×      / |
/__|         /__|
3,4,5       9,12,15

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4. Triangle Congruency

• Formula: Two triangles are congruent if they satisfy standard conditions:

$$\boxed{SSS,\ SAS,\ ASA,\ AAS,\ RHS}$$

Corresponding sides and angles of congruent triangles are equal. Use when: The question asks whether two triangles are exactly equal in shape and size.

• Example: Two right triangles have equal hypotenuse (10) cm and one corresponding side (6) cm. Are they congruent?

Solution: Both are right triangles.

They have:

• Equal hypotenuse (=10) cm
• Equal corresponding side (=6) cm
• Equal right angle (=90^\circ)

Therefore, by RHS congruency:

$$\boxed{\text{The triangles are congruent}}$$

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5. Polygon Interior & Exterior Angles

• Formula:

$$\boxed{\text{Sum of interior angles}=(n-2)180^\circ}$$

For a regular polygon:

$$\boxed{\text{Each interior angle}=\frac{(n-2)180^\circ}{n}}$$ $$\boxed{\text{Each exterior angle}=\frac{360^\circ}{n}}$$

Use when: Finding the number of sides or individual angles of a polygon.

• Example: The sum of interior angles of a polygon is (1260^\circ). Find the number of sides.

Solution:

$$(n-2)180=1260$$ $$n-2=7$$ $$\boxed{n=9}$$

The polygon has 9 sides.

Interior angle sum:
(n - 2) × 180°
          ↓
       1260°
          ↓
         n = 9

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6. Regular Polygon Diagonals

• Formula:

$$\boxed{\text{Number of diagonals}=\frac{n(n-3)}2}$$

Use when: A polygon’s number of sides is known and the number of diagonals is required.

• Example: How many diagonals does a decagon have?

Solution: For (n=10):

$$D=\frac{10(10-3)}2$$ $$=\frac{10\times7}{2}$$ $$\boxed{35}$$

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7. Circle Chord, Tangent & Radius Properties

• Formula:

$$\boxed{\text{Radius}\perp\text{tangent at point of contact}}$$

Tangents drawn from the same external point are equal:

$$\boxed{PA=PB}$$

Use when: A tangent touches a circle or two tangents are drawn from the same external point.

• Example: From an external point (P), two tangents touch a circle at (A) and (B). If (PA=12) cm, find (PB).

Solution: Tangents from the same external point are equal:

$$PA=PB$$

Therefore:

$$\boxed{PB=12\text{ cm}}$$

          A
         /|
        / |
       /  | radius
      P---O
       \  |
        \ |
         \|
          B

PA = PB

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8. Cyclic Quadrilateral Angles

• Formula: Opposite angles of a cyclic quadrilateral are supplementary:

$$\boxed{\angle A+\angle C=180^\circ}$$ $$\boxed{\angle B+\angle D=180^\circ}$$

Also, the angle subtended by the same chord is equal. Use when: All four vertices lie on the same circle.

• Example: (ABCD) is cyclic and (\angle A=75^\circ). Find (\angle C).

Solution: Opposite angles are supplementary:

$$\angle A+\angle C=180^\circ$$ $$75^\circ+\angle C=180^\circ$$ $$\boxed{\angle C=105^\circ}$$

        A──────B
      /          \
    D              C
      \__________/
      
∠A + ∠C = 180°

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9. Circumradius & Inradius

• Formula:

$$\boxed{R=\frac{a}{2\sin A}}$$

where (R) is the circumradius.

$$\boxed{r=\frac{\Delta}{s}}$$

where (r) is the inradius, (\Delta) is area, and

$$s=\frac{a+b+c}{2}.$$

Use when: A triangle’s sides/angles and its inscribed or circumscribed circle are involved.

• Example: Find the circumradius of a triangle having side (a=10) cm opposite (30^\circ).

Solution:

$$R=\frac{a}{2\sin A}$$ $$R=\frac{10}{2\sin30^\circ}$$

Since (\sin30^\circ=\frac12):

$$R=\frac{10}{1}=10$$ $$\boxed{R=10\text{ cm}}$$

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10. Equilateral Triangle & Inscribed/Circumscribed Circles

• Formula: For an equilateral triangle of side (a):

$$\boxed{R=\frac{a}{\sqrt3}},\qquad \boxed{r=\frac{a}{2\sqrt3}}$$

Hence,

$$\boxed{R:r=2:1}$$

and

$$\boxed{\text{Circumcircle area : Incircle area}=4:1}.$$

Use when: An equilateral triangle has both an incircle and circumcircle.

• Example: An equilateral triangle has side (6) cm. Find the ratio of the circumcircle area to the incircle area.

Solution:

$$R=\frac6{\sqrt3}=2\sqrt3$$ $$r=\frac6{2\sqrt3}=\sqrt3$$

Therefore:

$$\frac{\pi R^2}{\pi r^2} =\frac{(2\sqrt3)^2}{(\sqrt3)^2} =\frac{12}{3}$$ $$\boxed{4:1}$$

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11. Basic Circle Angle Theorems

• Formula:

$$\boxed{\text{Angle at centre}=2\times\text{angle at circumference}}$$

for the same arc.

Angle in a semicircle:

$$\boxed{90^\circ}$$

Use when: Central angles and angles formed on the circumference subtend the same arc.

• Example: An angle at the circumference subtending an arc is (35^\circ). Find the corresponding central angle.

Solution:

$$\text{Central angle}=2(35^\circ)$$ $$\boxed{70^\circ}$$

          A
        /   \
       /     \
      O-------B
       \     /
        \___/

∠AOB = 2∠ACB

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Advanced Variants

12. Incenter Angle Property

• Formula: If (I) is the incenter of (\triangle ABC), then

$$\boxed{\angle BIC=90^\circ+\frac{\angle A}{2}}$$

Similarly,

$$\angle CIA=90^\circ+\frac{\angle B}{2}.$$

Use when: A triangle contains its incenter and an angle involving two angle bisectors is required.

• Example: In (\triangle ABC), (\angle A=60^\circ). If (I) is the incenter, find (\angle BIC).

Solution:

$$\angle BIC=90^\circ+\frac{60^\circ}{2}$$ $$=90^\circ+30^\circ$$ $$\boxed{120^\circ}$$

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13. Area Ratio of Similar Triangles

• Formula: If corresponding sides are in ratio (m:n), then

$$\boxed{\text{Area ratio}=m^2:n^2}.$$

Use when: Similar triangles have known side or area ratios.

• Example: Two similar triangles have corresponding sides in ratio (2:3). Find their area ratio.

Solution:

$$\text{Area ratio}=2^2:3^2$$ $$\boxed{4:9}$$

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14. Tangent-Secant / Intersecting Chord Relations

• Formula: For tangent (PT) and secant (PAB):

$$\boxed{PT^2=PA\times PB}$$

For two intersecting chords:

$$\boxed{PA\times PB=PC\times PD}.$$

Use when: A circle problem contains a tangent or two chords intersecting inside the circle.

• Example: A tangent from (P) has length (12) cm. A secant from (P) meets the circle at distances (9) cm and (x) cm. Find (x).

Solution:

$$PT^2=PA\times PB$$ $$12^2=9x$$ $$144=9x$$ $$\boxed{x=16\text{ cm}}$$

             T
            / 
           / 12
          /
P────────A────────B
     9       x

PT² = PA × PB