In the triangle ABC, the points D and E are on AB and AC, and the angle EBC = angle DCB = I / 2 angle a, BD = CE is proved

Suppose the angle a = 2 * x, CD and be intersect at O, then
1. In the triangle Coe, the angle COE = 2 * x, the angle CEO = B + X, according to the sine theorem
CE/sin(2*x)=CO/sin(B+x)
2. In the triangle BOD, the angle BOD = 2 * x, the angle BDO = Pi - (2 * x + b-X) = Pi - (B + x)
BD/sin(2*x)=BO/sin(B+x)
3. Angle EBC = angle DCB = = > Bo = Co
So finally, BD = CE

In the triangle ABC, the points D and E are on AB and AC, and the angle EBC = angle DCB = I / 2 angle a, BD = CE is proved

In the triangle ABC, the points D and E are on AB and AC, and the angle EBC = angle DCB = I / 2 angle a, BD = CE is proved

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