Question

ABCD is a rectangle with BC = a, AB = b and a > b. If M is a point on AD such that ∠BMA = ∠BMC, then MD is equal to:

(a) \(\sqrt{a^2+b^2}\)

(b) \(\sqrt{a^2-b^2}\)

(c) \(\sqrt{ab}\) 

(d) a – b

Answer 1

 (b) \(\sqrt{a^2-b^2}\)

Let ∠BMA = ∠BMC = x 

Then ∠CMD = π – 2x 

Let MD = y. Then AM = (a – y) 

In ∆CMD, 

tan (π – 2x) = \(\frac{b}{y}\)                    ...(i) 

In ∆BAM, tan x = \(\frac{b}{a-y}\)           ...(ii) 

From (i), we have tan (π – 2x) = – tan 2x = \(\frac{b}{y}\)

⇒ – 2bya + 2by² = ba² + by² – 2aby – b³ 

⇒ by² = b(a² – b²)

⇒ y = \(\sqrt{a^2-b^2}\)