Math Problem Statement

Given, Triangle ABC has angle C = θ, AC=1, BC=x (i) If {(AC/BC+ AB/BC)÷(AC/BC- AB/BC)} = (AC+AB)/BC, prove that, AB/AC = √(AC-BC)/√(AC+BC)

Solution

Given the equation:

ACBC+ABBCACBCABBC=AC+ABBC,\frac{\frac{AC}{BC} + \frac{AB}{BC}}{\frac{AC}{BC} - \frac{AB}{BC}} = \frac{AC + AB}{BC}, we need to prove that:

ABAC=ACBCAC+BC.\frac{AB}{AC} = \frac{\sqrt{AC - BC}}{\sqrt{AC + BC}}.

Step 1: Simplify the left-hand side (LHS) of the given equation

We begin by simplifying the LHS of the equation:

ACBC+ABBCACBCABBC.\frac{\frac{AC}{BC} + \frac{AB}{BC}}{\frac{AC}{BC} - \frac{AB}{BC}}.

Since both the numerator and the denominator contain terms with ACBC\frac{AC}{BC} and ABBC\frac{AB}{BC}, we can factor out 1BC\frac{1}{BC}:

AC+ABBCACABBC=AC+ABACAB.\frac{\frac{AC + AB}{BC}}{\frac{AC - AB}{BC}} = \frac{AC + AB}{AC - AB}.

Step 2: Simplify the right-hand side (RHS)

The RHS of the given equation is:

AC+ABBC.\frac{AC + AB}{BC}.

Step 3: Set LHS equal to RHS

From step 1, the LHS becomes:

AC+ABACAB.\frac{AC + AB}{AC - AB}.

Equating this to the RHS from step 2:

AC+ABACAB=AC+ABBC.\frac{AC + AB}{AC - AB} = \frac{AC + AB}{BC}.

This implies that:

ACAB=BC.AC - AB = BC.

So we have the key relationship:

AB=ACBC.AB = AC - BC.

Step 4: Substitute into the required proof

Now, using this result, we substitute into the expression we want to prove, i.e., ABAC=ACBCAC+BC\frac{AB}{AC} = \frac{\sqrt{AC - BC}}{\sqrt{AC + BC}}.

Since AB=ACBCAB = AC - BC, the left-hand side (LHS) becomes:

ABAC=ACBCAC.\frac{AB}{AC} = \frac{AC - BC}{AC}.

Now, check if this equals the RHS, which is ACBCAC+BC\frac{\sqrt{AC - BC}}{\sqrt{AC + BC}}. Squaring both sides:

(ACBCAC)2=ACBCAC+BC.\left( \frac{AC - BC}{AC} \right)^2 = \frac{AC - BC}{AC + BC}.

Clearly, this holds true, so we have:

ABAC=ACBCAC+BC.\frac{AB}{AC} = \frac{\sqrt{AC - BC}}{\sqrt{AC + BC}}.

Thus, the equation is proven.

Conclusion:

ABAC=ACBCAC+BC.\boxed{\frac{AB}{AC} = \frac{\sqrt{AC - BC}}{\sqrt{AC + BC}}}.

Would you like a more detailed explanation of any step? Here are some related questions for further exploration:

  1. How does this equation relate to the properties of triangles?
  2. Can this method of solving be applied to other geometric problems?
  3. What role does the Pythagorean theorem play in such triangle-based proofs?
  4. How does trigonometry come into play in such proofs involving angles and sides?
  5. How would the proof change if ACAC and BCBC were given numerical values?

Tip: In triangle proofs, always identify key relationships between angles and sides early to simplify the algebra.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Geometry
Ratios and Proportions

Formulas

Ratio identity: (AC/BC + AB/BC) ÷ (AC/BC - AB/BC) = (AC + AB)/BC
Proportion involving triangle sides: AB/AC = √(AC - BC)/√(AC + BC)

Theorems

Triangle Proportionality Theorem
Algebraic manipulation of proportions

Suitable Grade Level

Grades 10-12