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 Triangle right angled at B and BD perpendicular to AC .

To prove:-

• BD²=AD.CD

Proof:-

In triangle ABC, BD ⟂ AC

show /\ BDC ~ ADB (∵perpendicular is drawn from the right vertex of a right triangle then the Triangles from either side of the perpendicular are similar to each other and similar to the whole Triangle)

$⟹so \: \frac{BD}{AD} = \frac{CD}{BD}$

⟹BD.BD = AD.CD

⟹BD² = AD.CD.

Hence Proved.

Answer 2

Step-by-step explanation:

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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