解析 解:∵ ADperp;DB,there4;ang;ADB=90deg;. ∵ ang;ACD=70deg;,there4;ang;DAC=20deg;. ∵ ang;B=30deg;,there4;ang;DAB=60deg;, there4;ang;CAB=40deg;. ∵ AE平分ang;CAB, there4;ang;BAE=20deg;, there4; ang;AED=ang;BAE +ang;B =50deg;....
已知,点M、N分别是AB、CD上两点,点G在AB、CD之间,连接MG、NG.(1)如图1,若GM⊥ GN,求∠ ANG+∠ CNG的度数;(2)如图2,若点P是CD下方
已知:如图,\(AC\bot BC\),垂足为\(C\),\( \angle BCD\)是\( \angle B\)的余角.求证:\( \angle ACD = \angle B\).证明:\( \because AC\bot BC\)(已知),\( \therefore \angle ACB = 90^{\circ}\)(___).\( \therefore \angle BCD\)是\( \angle ACD\)的余角.\( \because...
如图,在△ABC中,AB =AC,AD是BC边上的中线,四边形 ADBE 是平行四边形,求证:四边形ADBE是矩形.E Ang B DC