Question

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Answer 1

Question:

AD is an altitude of an isosceles triangle in which AB = AC. Show that

i) AD bisects BC  ii) AD bisects ∠A

Answer:

$\Large{\underline{\underline{\bf{Given:}}}}$

• AB = AC
• AD is an altitude of the isosceles triangle

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

• AD bisects BC
• AD bisects ∠A

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Consider Δ ABD and Δ ACD

  ∠ADB = ∠ADC = 90° (altitude maked an angle of 90°)

  AB = AC (given)

 ∠ABD = ∠ACD (Base angles are equal in an isosceles triangle)

Hence Δ ABD ≅ Δ ACD

→ Therefore BD = CD (by CPCT)

  ∴ AD bisects BC

→ ∠BAD = ∠DAC (by CPCT)

  ∴ AD bisects ∠A

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ In an isoceles triangle, base angles are equal.

→ In an equilateral triangle measure of all angles = 60°

Answer 2

dekh le...................