已知:如图,在△ABC中,AB=AC,AD⊥BC,垂足为点D,AN是△ABC外角∠CAM的平分线,CE⊥AN,垂足为点E,

2025-05-05 02:17:03
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回答1:

(1)证明:在△ABC中, AB=AC,AD⊥BC.
                    ∴ ∠BAD=∠DAC. ∵ AN是△ABC外角∠CAM的平分线,
                      ∴ . ∴ ∠DAE=∠DAC+∠CAE= 180°=90°.
                    又 ∵ AD⊥BC,CE⊥AN, ∴ =90°,
                    ∴ 四边形ADCE为矩形;
(2)当AD= 时,四边形ADCE是正方形.
         证明:∵ AB=AC,AD⊥BC于D.
                    ∴ DC= . 
                  又 AD= ,∴ DC=AD.
            由(1)四边形ADCE为矩形,
              ∴ 矩形ADCE是正方形.