试题分析:(1)以点N为圆心,以MQ长度为半径画弧,以点M为圆心,以NQ长度为半径画弧,两弧交于一点F,则△MNF为所画三角形.
(2)延长DA至E,使得AE=CB,连结CE.证明△EAC≌△BCA,得:∠B =∠E,AB=CE,根据等量代换可以求得答案.
试题解析:(1)如图1,以N 为圆心,以MQ 为半径画圆弧;以M 为圆心,以NQ 为半径画圆弧;两圆弧的交点即为所求.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082303391521414136.png)
(2)如图,延长DA至E,使得AE=CB,连结CE.
∵∠ACB +∠CAD =180°,∠DACDAC +∠EAC =180°,∴∠BACBCA =∠EAC.
在△EAC和△BAC中,AE=CE,AC=CA,∠EAC=∠BCN,
∴△AECEAC≌△BCA (SAS).∴∠B=∠E,AB=CE.
∵∠B=∠D,∴∠D=∠E.∴CD=CE,∴CD=AB.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082303391523014077.png)