prove that angle abc is an isosceles triangle if the altitude ABC from A on BC bisects BC
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In ∆ADB and ∆ADC AD = AD [Common side]
∠APB =∠ ADC [90°] BD = DC [AD bisects BC]
∴∆ADB ≅ ∆ADC [SAS postulate]
∴AB = AC [Corresponding sides]
∴∆ ABC is an isosceles triangle
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