Interactive Geometry Exercises

Master geometry with step-by-step solutions and interactive diagrams!

1 Triangle Similarity Proof

In the figure, if \( BD \perp AC \) and \( CE \perp AB \), prove that:
(i) \( \triangle AEC \sim \triangle ADB \)
(ii) \( \frac{CA}{AB} = \frac{CE}{DB} \)

Using AA (Angle-Angle) Similarity Criterion

Given: \( BD \perp AC \) and \( CE \perp AB \)

∴ ∠ADB = ∠AEC = 90°

For part (i): Prove \( \triangle AEC \sim \triangle ADB \)

In ΔAEC and ΔADB:

1. ∠AEC = ∠ADB = 90° (given)

2. ∠EAC = ∠DAB (common angle)

Therefore, by AA similarity criterion:

ΔAEC ∼ ΔADB

For part (ii): Prove \( \frac{CA}{AB} = \frac{CE}{DB} \)

From the similarity ΔAEC ∼ ΔADB, corresponding sides are proportional:

\(\frac{AE}{AD} = \frac{AC}{AB} = \frac{EC}{DB}\)

Therefore:

\(\frac{CA}{AB} = \frac{CE}{DB}\)

Both parts of the problem are now proved using triangle similarity!

2 Parallel Lines Problem

In the given figure \( AB \parallel CD \parallel EF \).
If \( AB = 6 \, \text{cm}, \, CD = x \, \text{cm}, \, EF = 4 \, \text{cm}, \, BD = 5 \, \text{cm} \) and \( DE = y \, \text{cm} \). Find \( x \) and \( y \).

Using Basic Proportionality Theorem (Thales' Theorem)

Given three parallel lines \( AB \parallel CD \parallel EF \) with transversals.

Using the property of parallel lines and transversals:

\(\frac{AB}{CD} = \frac{BD}{DE}\)

\(\frac{6}{x} = \frac{5}{y}\) ...(1)

Also:

\(\frac{CD}{EF} = \frac{BD}{BE}\)

\(\frac{x}{4} = \frac{5}{5 + y}\) ...(2)

From equation (1): \( 6y = 5x \) ⇒ \( y = \frac{5x}{6} \)

Substitute into equation (2):

\(\frac{x}{4} = \frac{5}{5 + \frac{5x}{6}}\)

Solve for x:

\(\frac{x}{4} = \frac{5}{\frac{30 + 5x}{6}} = \frac{30}{30 + 5x}\)

x(30 + 5x) = 120

30x + 5x² = 120

5x² + 30x - 120 = 0

x² + 6x - 24 = 0

Using quadratic formula:

x = [-6 ± √(36 + 96)]/2 = [-6 ± √132]/2 = -3 ± √33

Taking positive value (since length is positive):

x ≈ 2.744 cm

Then y = \( \frac{5 \times 2.744}{6} ≈ 2.287 \) cm

The solutions are: \( x ≈ 2.744 \, \text{cm} \) and \( y ≈ 2.287 \, \text{cm} \)

3 Ceva's Theorem Application

O is any point inside a triangle \( ABC \). The bisector of \( \angle AOB \), \( \angle BOC \) and \( \angle COA \) meet the sides \( AB, BC \) and \( CA \) in point \( D, E \) and \( F \) respectively. Show that \( AD \times BE \times CF = DB \times EC \times FA \)

Using Angle Bisector Theorem and Ceva's Theorem

Using Angle Bisector Theorem repeatedly:

In ΔAOB, OD bisects ∠AOB:

\(\frac{AD}{DB} = \frac{OA}{OB}\) ...(1)

In ΔBOC, OE bisects ∠BOC:

\(\frac{BE}{EC} = \frac{OB}{OC}\) ...(2)

In ΔCOA, OF bisects ∠COA:

\(\frac{CF}{FA} = \frac{OC}{OA}\) ...(3)

Multiplying equations (1), (2), and (3):

\(\frac{AD}{DB} \times \frac{BE}{EC} \times \frac{CF}{FA} = \frac{OA}{OB} \times \frac{OB}{OC} \times \frac{OC}{OA} = 1\)

Therefore:

\(AD \times BE \times CF = DB \times EC \times FA\) (QED)

The required product relationship is proved using angle bisector properties.

4 Cyclic Quadrilateral Proof

In the figure, \( ABC \) is a triangle in which \( AB = AC \). Points \( D \) and \( E \) are points on the side \( AB \) and \( AC \) respectively such that \( AD = AE \). Show that the points \( B, C, E \) and \( D \) lie on a same circle.

Using Properties of Isosceles Triangles and Cyclic Quadrilaterals

Given AB = AC and AD = AE ⇒ AB - AD = AC - AE ⇒ DB = EC

Consider ΔADE: AD = AE ⇒ ΔADE is isosceles ⇒ ∠ADE = ∠AED

∠ADE + ∠BDE = 180° (linear pair)

∠AED + ∠CED = 180° (linear pair)

But ∠ADE = ∠AED ⇒ ∠BDE = ∠CED

In ΔBDE and ΔCED:

1. BD = CE (from step 1)

2. DE is common

3. ∠BDE = ∠CED (from step 3)

∴ ΔBDE ≅ ΔCED (by SAS congruency)

Thus, ∠DBE = ∠ECD

In quadrilateral BCED:

∠DBE + ∠DCE = ∠ECD + ∠DCE = ∠ACB + ∠DCE = ∠ECB

But AB = AC ⇒ ∠ABC = ∠ACB

Thus, ∠DBE + ∠DCE = ∠ABC

But ∠ABC + ∠ECB = 180° (angles on straight line)

∴ ∠DBE + ∠DCE + ∠ECB = 180°

The sum of opposite angles of quadrilateral BCED is 180°, hence it is cyclic (QED)

5 Moving Trains Problem

Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels at a speed of 20 km/hr and the second train travels at 30 km/hr. After 2 hours, what is the distance between them?

Using Pythagoras' Theorem

Calculate distance traveled by each train in 2 hours:

First train (west): \( 20 \, \text{km/hr} \times 2 \, \text{hr} = 40 \, \text{km} \)

Second train (north): \( 30 \, \text{km/hr} \times 2 \, \text{hr} = 60 \, \text{km} \)

The paths form a right-angled triangle with legs 40 km and 60 km.

Use Pythagoras' theorem to find the distance between them:

\( D = \sqrt{40^2 + 60^2} = \sqrt{1600 + 3600} = \sqrt{5200} \)

\( = \sqrt{100 \times 52} = 10 \times \sqrt{52} \)

\( ≈ 72.11 \, \text{km} \)

After 2 hours, the distance between the trains is approximately 72.11 km.

6 Triangle Relations Proof

\( D \) is the mid point of side \( BC \) and \( AE \perp BC \). If \( BC = a, AC = b, AB = c, ED = x \), \( AD = p \) and \( AE = h \), prove that:
(i) \( b^2 = p^2 + ax + \frac{a^2}{4} \)
(ii) \( c^2 = p^2 - ax + \frac{a^2}{4} \)
(iii) \( b^2 + c^2 = 2p^2 + \frac{a^2}{2} \)

Using Pythagoras' Theorem in Right Triangles

Given: D is midpoint ⇒ BD = DC = a/2

AE ⊥ BC ⇒ ∠AEB = ∠AEC = 90°

For part (i): Prove \( b^2 = p^2 + ax + \frac{a^2}{4} \)

In right ΔAEC:

AC² = AE² + EC² ⇒ b² = h² + (ED + DC)²

b² = h² + (x + a/2)² = h² + x² + ax + a²/4

In right ΔAED:

AD² = AE² + ED² ⇒ p² = h² + x² ⇒ h² + x² = p²

Substitute into part (i) result:

b² = p² + ax + a²/4 (QED)

For part (ii): Prove \( c^2 = p^2 - ax + \frac{a^2}{4} \)

In right ΔAEB:

AB² = AE² + EB² ⇒ c² = h² + (BD - ED)²

c² = h² + (a/2 - x)² = h² + a²/4 - ax + x²

Using h² + x² = p²:

c² = p² - ax + a²/4 (QED)

For part (iii): Prove \( b^2 + c^2 = 2p^2 + \frac{a^2}{2} \)

Add results from (i) and (ii):

b² + c² = (p² + ax + a²/4) + (p² - ax + a²/4)

= 2p² + a²/2 (QED)

All three relations are now proved using Pythagoras' theorem.

7 Mirror Height Problem

A man whose eye-level is 2 m above the ground wishes to find the height of a tree. He places a mirror horizontally on the ground 20 m from the tree and finds that if he stands at a point C which is 4 m from the mirror B, he can see the reflection of the top of the tree. How tall is the tree?

Using Properties of Similar Triangles and Reflection

The mirror creates similar triangles through reflection:

\(\frac{\text{Eye height}}{\text{Distance to mirror}} = \frac{\text{Tree height}}{\text{Distance from mirror to tree}}\)

Given:

Eye height = 2 m

Distance to mirror = 4 m

Distance from mirror to tree = 20 m

Let tree height = h

Set up the proportion:

\(\frac{2}{4} = \frac{h}{20}\)

Solve for h:

\( h = \frac{2 \times 20}{4} = 10 \, \text{m} \)

The tree is 10 meters tall.

8 Emu Shadow Problem

An Emu which is 3 feet tall is standing at the foot of a pillar which is 30 feet high. It walks away from the pillar. The shadow of the Emu falls beyond Emu. What is the relation between the length of the shadow and the distance from the Emu to the pillar?

Using Properties of Similar Triangles

Let:

s = length of emu's shadow

d = distance from emu to pillar

The triangles formed are similar (AA similarity):

\(\frac{\text{Emu height}}{\text{Pillar height}} = \frac{\text{Shadow length}}{\text{Total distance}}\)

\(\frac{3}{30} = \frac{s}{d + s}\)

Simplify the proportion:

\(\frac{1}{10} = \frac{s}{d + s}\)

Cross-multiply:

d + s = 10s

d = 9s

The relation between shadow length and distance is: \( d = 9s \) or \( s = \frac{d}{9} \)

9 Circle Geometry Proof

Two circles intersect at \( A \) and \( B \). From a point \( P \) on one of the circles lines \( PAC \) and \( PBD \) are drawn intersecting the second circle at \( C \) and \( D \). Prove that \( CD \) is parallel to the tangent at \( P \).

Using Alternate Segment Theorem and Cyclic Quadrilaterals

Let PT be the tangent at P to the first circle.

By alternate segment theorem:

∠PTA = ∠PBA ...(1)

In the second circle, quadrilateral ABDC is cyclic:

∠PBA = ∠PCD ...(2) (exterior angle equals interior opposite angle)

From (1) and (2):

∠PTA = ∠PCD

These are corresponding angles formed by transversal PC with PT and CD.

Therefore, PT ∥ CD (QED)

10 Triangle Section Problem

Let \( ABC \) be a triangle and \( D,E,F \) are points on the respective sides \( AB \), \( BC \), \( AC \) (or their extensions). Let \( AD:DB = 5:3 \), \( BE:EC = 3:2 \) and \( AC = 21 \). Find the length of the line segment \( CF \).

Using Menelaus' Theorem

Given ratios:

AD/DB = 5/3 ⇒ BD/DA = 3/5

BE/EC = 3/2 ⇒ CE/EB = 2/3

AC = 21

Apply Menelaus' Theorem to ΔABC with transversal DEF:

\(\frac{AF}{FC} \times \frac{CE}{EB} \times \frac{BD}{DA} = 1\)

Let AF = x ⇒ FC = 21 - x

Substitute known ratios:

\(\frac{x}{21 - x} \times \frac{2}{3} \times \frac{3}{5} = 1\)

Simplify:

\(\frac{x}{21 - x} \times \frac{6}{15} = 1\)

\(\frac{x}{21 - x} \times \frac{2}{5} = 1\)

Cross-multiply:

2x = 5(21 - x)

2x = 105 - 5x

7x = 105

x = 15

Therefore, CF = AC - AF = 21 - 15 = 6 units.