Math, asked by aayushsingh2614, 11 months ago

In an isosceles triangle ABC with AB = AC, BD is perpendicular from B to the
side AC. Prove that
BD2-CD2=2CD.AD

Answers

Answered by AditiHegde
0

In an isosceles triangle ABC with AB = AC, BD is perpendicular from B to the

side AC.

Consider the attached figure while going through the following steps.

ABC is an isosceles triangle

BD ⊥ AC

Now consider, in Δ ABD,

AB² = AD² + BD²

AB = AC (sides of an isosceles triangle are equal)

⇒ AC² = AD² + BD²

we have, AC = AD + DC

⇒ (AD + DC)² = AD² + BD²

⇒ AD² + DC² + 2 × AD × DC = AD² + BD²

⇒ DC² + 2 × AD × DC = BD²

⇒ 2 × AD × DC = BD² - DC²

BD² - DC² = 2 CD.AD

Hence proved.

Attachments:
Answered by premppk
0

Answer:

Step-by-step explanation:

In an isosceles triangle ABC with AB = AC, BD is perpendicular from B to the

side AC.

Consider the attached figure while going through the following steps.

ABC is an isosceles triangle

BD ⊥ AC

Now consider, in Δ ABD,

AB² = AD² + BD²

AB = AC (sides of an isosceles triangle are equal)

⇒ AC² = AD² + BD²

we have, AC = AD + DC

⇒ (AD + DC)² = AD² + BD²

⇒ AD² + DC² + 2 × AD × DC = AD² + BD²

⇒ DC² + 2 × AD × DC = BD²

⇒ 2 × AD × DC = BD² - DC²

∴ BD² - DC² = 2 CD.AD

Hence proved.

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