In the adjoining figure, AB = AC and BD = DC. Prove that ΔADB ≅ ΔADC and hence show that
(i) ∠ADB = ∠ADC = 90°, (ii) ∠BAD = ∠CAD.

In ADB and ADCAB = AC givenBD = DC   given         AD = AD  commonADB  ADC  SSSADB = ADC  CPCTNow, ADB + ADC = 180°  linear pairADB  + ADB  = 180°ADB = 90° = ADCAlso, BAD = CAD   CPCTABC is an isosceles .

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turn the triangle up side then u can see a straight line that is 180. then subtract 180-90=90 
therefore u get triangle adb=90
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