问道数学题,快,在三角形ABC中,AB=AC,p是三角形内的一点且有角APB大于角APC.求证：PB小于PC

cos角APB = (AP^2 + BP^2 - AB^2)/(2AP*BP)
cos角APC = (AP^2 + CP^2 - AC^2)/(2AP*CP)
角APB > 角APC ==> cos角APB < cos角APC
==> (AP^2 + BP^2 - AB^2)/(2AP*BP) < (AP^2 + CP^2 - AB^2)/(2AP*CP) (AB=AC)
==> (AP^2 + BP^2 - AB^2)CP - (AP^2 + CP^2 - AB^2)BP < 0
==> AP^2 * (CP - BP) + BP*CP(BP - CP) - AB^2 * (CP - BP) < 0
==> (CP - BP)(AP^2 - BP*CP - AB^2) < 0
==> (CP - BP)(AB^2 + BP*CP - AP^2) > 0
AB > AP ==> AB^2 + BP*CP - AP^2 > 0
==> CP - BP > 0, BP < CP

很简单，PP'C大于P'PC，大边对大角，得证
你和我在线交流吧

当p在三角形abc的中线上时角apb=角apc
因为p离ab越近角apb越大
所以p在三角形abc中线的左边
做pk垂直于bc，此时bk小于kc，由勾股定理知pb小于pc