Question

一道初二数学题,急啊..............如图.在△ABC中,AB=AC,<BAD=20°.且AE=AD,求<CDE的度数.

Answer 1

10°
设∠B度数为X，AB=AC。∠C也为X
∠DAE=180-2X-20
因为AD=AE，∠AED=（180-∠DAE）/2=X+10
∠AED是三角形ECD的外角，∠AED=∠CDE+∠C
即∠CDE+X=X+10，所以∠CDE=10

Answer 2

∵∠ADC=∠B+∠BAD=20°+∠B,∠ADE=∠AED=∠C+∠CDE
∠ADE=∠AED,∠B=∠C
∴20°+∠B=∠ADE+∠CDE=∠C+∠CDE+∠CDE
∴2∠CDE=20°
∴∠CDE=10°

Answer 3

解：令∠BAC=x° 则∠B=（180-x）/2 ∠DAE=x-20 ∠ADE=(180-X+20)/2
=100-x/2
∠ADC=∠B+∠BAD=（180-x）/2 +20=110-x/2
∴∠CDE=∠ADC-∠ADE=110-x/2- 100+x/2=10°