ABC is an isosceles triangle with a b is equals to easy draw a perpendicular BC and show that AngleB =AngleC
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Given: in ΔABC, AB = AC
To prove: ∠B = ∠C
Proof: Consider Δ’s ADB and ADC
AB = AC (Given)
∠ADB = ∠ADC = 90° [Since AD ⊥ BC]
AD = AD [Common side]
⇒ ΔADB ≅ ΔADC [By RHS congruence rule]
∴ ∠B = ∠C [CPCT]
To prove: ∠B = ∠C
Proof: Consider Δ’s ADB and ADC
AB = AC (Given)
∠ADB = ∠ADC = 90° [Since AD ⊥ BC]
AD = AD [Common side]
⇒ ΔADB ≅ ΔADC [By RHS congruence rule]
∴ ∠B = ∠C [CPCT]
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