Solving Geometry Problems: A Step-by-Step Guide

Geometry is a branch of mathematics that deals with shapes, sizes, and properties of space. Solving geometry problems involves understanding various properties of geometric shapes and relationships between angles and lines. In this guide, we will demonstrate a step-by-step approach to solving a classic geometry problem.

Problem Context and Notations

Let's symbolize the vertex of the right angle as T. Consider the given scenario in a geometric diagram where we need to solve for certain angles and prove specific relationships.

Step 1: Identifying Similar Triangles

Triangles △RTB and △RTC have:

Angle ∠TRB ∠TRC Common line segment RT Angle ∠RTB ∠RTC

Therefore, by the Angle-Angle-Side (AAS) congruence theorem, △RTB ? △RTC. As a result, we get:

Angle ∠RCT Angle ∠RBT m Angle ∠QCD Angle ∠RCT because they are vertical angles

Step 2: Using Triangle Angle Sum Theorem

On triangle △RTB, we have:

Angle m 90° θ 180°

Solving for θ, we get:

θ 90° - m

Step 3: Utilizing Externals Angles

Angle ∠PSR is the external angle of △RSQ and can be expressed as:

Angle ∠PSR θ Angle ∠PSR Angle ∠θP 180°

Solving for qp and θ, we get:

θqθp 180° qp 180° - 2θ qp 2(90° - m) qp 180° - 2m qp 2m

Thus, m qp / 2.

Step 4: Analyzing More Triangles

Angle ∠SQR is the external angle of △QCD, which gives us:

Angle ∠SQR Angle ∠q Angle ∠SQR Angle qp / 2 (which implies that d qp / 2)

Step 5: Proving Isosceles Triangles

Triangles △BAE and △CDE have:

Length BA Length CD Angle ∠BAE Angle ∠CDE 75° Length AE Length ED

By the Side-Angle-Side (SAS) congruence theorem, △BAE ? △CDE.

Step 6: Extending Lines and Forming New Triangles

Extend line AE until it meets CD at point F. Then, ∠AFD 75°, and ∠EFD 75°. Hence, △EFD is isosceles, meaning EF ED.

Angle ∠FED is the external angle of △EAD and equals 30°.

Step 7: Auxiliary Lines and Midpoints

Form the diagonal AC. From point F draw a perpendicular to AC which meets AC at point K.

This gives us:

Angle ∠KAF 30° Angle ∠KFA 60°

Step 8: Median Properties in Right Triangles

From point F, draw the line segment KE. Since KE is the median to the hypotenuse AF of right △KAF and point E is the midpoint of the hypotenuse, KE EA EF.

Since KE EA, △KEA is isosceles, and ∠KAE ∠AKE.

Angle ∠KEF is the external angle of △KEA and equals 30°.

This implies that △KEF is equilateral, so KF KE EF.

Angle ∠KFC 45°, and ∠KCF 45°, making △KCF isosceles with KC KF KE.

Step 9: Defining the Circumcircle

Point K is the center of the circumcircle of △CFE. Angle ∠KED 90°, meaning ED is tangent to the circumcircle of △CFE at point E.

Using the Alternate Segment Theorem, ∠ECD ∠FED 30°.

Step 10: Final Solution

Therefore, ∠EBA ∠ECD 30°. Angle ∠BCE ∠BCD - ∠ECD 60°, and ∠CBE ∠CBA - ∠EBA 60°. Since ∠BCE ∠CBE 60°, △CBE is an equilateral triangle.