Question

In a right angled triangle , if a perpendicular is drawn from right angle ti hypotenuse then prove that the area of the square of the perpendicular is equal to the area of the rectangle formed by the two segments of the hypotenuse .

Answer 1

Given: ABC is a right ∆ right angled at A

and AD ⊥ BC

Now in ∆ABC

∠A + ∠B + ∠C = 180° (Angle sum properly)

⇒∠B + ∠C = 180° – ∠A = 180° – 90° = 90° ........... (1)

In ∆ABD

∠B + ∠BAD + ∠ADB = 180°

⇒∠B + ∠BAD = 180° – ∠ADB = 180° – 90° = 90° ........... (2)

In ∆ADC

⇒∠DAC + ∠ADC + ∠C = 180°

⇒∠C + ∠DAC = 180° – ∠ADC = 180° – 90° = 90° ........... (3)

Now in ∆BAD and ∆ACD

∠BAD = ∠C (from (1) and (2))

∠B = ∠DAC (from (1) and (3))

∆BAD ~ ∆ACD (by AA similarly centers)

Thus BD : AD : : AD : CD

Hence AD is the mean proportional between BD and CD