Question

AB is an altitude of an isosceles triangle ABC is which AB=AC. Prove that <BAD = <DAC.
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Answer 1

Answer:

in ∆ ADB and ∆ ADC

AD=AD(common side)

Step-by-step explanation:

<ADB=<ADC(as AB altitude on side BC) AB=BC(given) therefore ∆ADB congruent ∆ADC(RHS) hence <BAD=<DAC(CPCT)