Question

in a right triangle if a perpendicular is drawn from the right angle to the hypotenuse, prove that the square of the perpendicular is equal to the rectangle contained 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


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