题目

如图，已知分别是的内角平分线，过点作；垂足分别为连结若则的长等于_______(用含的代数式表示结果)． 答案：【答案】()【解析】延长AG交BC于N，延长AF交BC于M，根据AF⊥BD，AG⊥CE，求证Rt△AGC≌Rt△NGC，可得AC=CN，AG=NG，同理可证：AF=FM，AB=BM．然后得出GF是△AMN的中位线，利用AB+AC=MB+CN=BN+MN+CM+MN，BC=BN+MN+CM，利用等量代换即可．延长AG交BC于N，延长AF交BC于M．∵AF⊥BD，AG⊥CE，∴∠AGC=∠CGN=90°，∠AFB=∠BFM=90°，在Rt△ Shelly had prepared carefully for her biology examination so that she could be sure of passing it at her first_________.        A. intention                 B. attempt                   C. purpose           D. desire