Question

ABC is an isosceles triangle whose <C is a right angle. if D is any point on A, prove that AD²+DB² = 2CD²​

Answer 1

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Answer 2

Figure in attachment .

[ Point D is on AB ]

Let AC = BC = x .

In Δ ADC and Δ BDC ,

DC = DC [ Common ]

AC = BC [ Given isosceles ]

∠ADC = ∠BDC [ 90° each ]

Δ ADC ≅ Δ BDC [ R.H.S ]

AD = DB [ c.p.c.t ]

AD² + DB² = AD² + AD²

⇒ 2 AD²

⇒ 2 ( AB/2 )²

⇒ 2 ( AB²/4 )

⇒ AB²/2

Now AB² = AC² + BC²

⇒ AB² = a² + a² ⇒  2 a²

Then AD² = ( AB/2 )² = AB²/4

⇒ 2 a²/4

⇒ a²/2

AB² = 2 a² / 2

⇒ AD² + DB² = a²

2 CD² = 2 (AC² - AD²)

2 CD² = 2 (a² - a²/2)

⇒ 2 CD² = 2 a²/2

⇒ 2 CD² = a²

Hence AD² + DB² = 2 CD² = a² [ Proved ]