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Question
AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that
- AD bisects BC
- AD bisects ∠A
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Solution
(i) In ∆ABD and ∆ACD,
AB = AC ...[Given]
∠ADB = ∠ADC ...[Each 90°]
AD = AD ...[Common]
∴ ∆ABD ≅ ∆ACD ...[By RHS Congruence Rule]
So, BD = DC ...[Corresponding parts of congruent triangles]
⇒ D is the mid-point of BC
or AD bisects BC.
(ii) Since, ∆ABD ≅ ∆ACD,
∠BAD = ∠CAD ...[Corresponding parts of congruent triangles]
⇒ Thus, AD bisects ∠A.
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