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Question:
AD is an altitude of an isosceles triangle in which AB = AC. Show that
i) AD bisects BC ii) AD bisects ∠A
Answer:
- AB = AC
- AD is an altitude of the isosceles triangle
- AD bisects BC
- AD bisects ∠A
→ 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
→ In an isoceles triangle, base angles are equal.
→ In an equilateral triangle measure of all angles = 60°
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