题目如图,在△ABC\triangle ABC△ABC中,点DDD在边BCBCBC上,且满足AB=AD=DCAB=AD=DCAB=AD=DC,过点DDD作DE⊥ADDE\bot ADDE⊥AD,交ACACAC于点EEE.设∠BAD=α\angle BAD=\alpha∠BAD=α,∠CAD=β\angle CAD=\beta∠CAD=β,∠CDE=γ\angle CDE=\gamma∠CDE=γ,则（ ）A.2α+3β=180∘2\alpha +3\beta =180^{\circ}2α+3β=180∘B.3α+2β=180∘3\alpha +2\beta =180^{\circ}3α+2β=180∘C.β+2γ=90∘\beta +2\gamma =90^{\circ}β+2γ=90∘D.2β+γ=90∘2\beta +\gamma =90^{\circ}2β+γ=90∘知识点：平行线的性质章节：图形的性质 / 相交线与平行线
如图,在中,点在边上,且满足,过点作,交于点.设,,,则（ ）如图,在中,点在边上,且满足,过点作,交于点.设,,,则（ ）如图,在中,点在边上,且满足,过点作,交于点.设,,,则（ ）