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Question
In a triangle ABC, if AD is the bisector of ∠BAC, which meets BC at point D such that AD ⊥ BC, then which of the following is always true?
In a triangle ABC, if AD is the bisector of ∠BAC, which meets BC at point D such that AD ⊥ BC, then which of the following is always true?
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Solution
The correct option is D ∠ABD = ∠ACD
Given that AD bisects ∠BAC and AD ⊥ BC.
In ΔABD and ΔACD,
∠BAD = ∠CAD (AD bisects ∠BAC)
AD = AD (Common)
∠ADB = ∠ADC (Each 90º)
∴ ΔABD ≅ ΔACD (ASA congruence rule)
⇒ ∠ABD = ∠ACD (CPCT)
Hence, the correct answer is option (d).
Given that AD bisects ∠BAC and AD ⊥ BC.
In ΔABD and ΔACD,
∠BAD = ∠CAD (AD bisects ∠BAC)
AD = AD (Common)
∠ADB = ∠ADC (Each 90º)
∴ ΔABD ≅ ΔACD (ASA congruence rule)
⇒ ∠ABD = ∠ACD (CPCT)
Hence, the correct answer is option (d).
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