Question

In triangle ABC, angle B=90°,BD perpendicular AC. AD=x,CD=y,BC=a,AB=c. Prove: 1/xy=1/c^2+ 1/a^2​

Answer 1

Given;

• In Δ ABC, ∠B=90°.
• Seg BD ⊥ Hypotenuse AC
• Seg AD= x ; Seg CD= y ; Side BC= a; Side AB=c.

To prove;

• $\frac{1}{xy} = \frac{1}{c^2} + \frac{1}{a^2}$  

Solution;

• ΔABC is right angled where ∠B= 90°.
• AC²= AB² + BC²....(1)  Using Pythagorus theorem

• We can find the area of triangle with two methods since we have two heights i.e BD & AB.
• ∴ A(ΔABC) = 1/2 × BD× AC = 1/2 ×AB× BC .... (ii)
• Substitute given values in above equation, we get
• 1/2 × BD × AC= 1/2 × a × c
• ∴ BD × AC = a × c
• On squaring, BD² × AC² = a² × c²... (iii)
• By property of geometric mean, BD²= AD × CD.
• BD²= x × y .... (iv)
• By property of collinearity, AC= x + y ...(v)
• From (iv) and (v) , eqn (iii) becomes
• (xy) × (x+y)² = a² × c².
• From (v) and (i),
• (x+y)² = a² + c²
• ∴ xy × (a² + c²) = a² × c².
• On taking inverse,
• 1/xy = (a² + c²) / a² × c²
• 1/xy= a²/a²c² + c²/a²c²   .......(Splitting numerator)
• 1/xy = 1/c² + 1/a².
• Hence, proved.

Answer;

1/xy = 1/c² + 1/a²