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