若PA=PB,角APB=2角ACB,AC与PB交于点D,且PB=4,PD=3,则AD*DC等于多少?
<p>问题：若PA=PB,角APB=2角ACB,AC与PB交于点D,且PB=4,PD=3,则AD*DC等于多少?<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">费定舟的回答：<div class="content-b">网友采纳 以P为圆心,以PA=PB为半径作圆,延长BD交圆于M, 则有：PA=PB=4,∠APB=2∠ACB,AC与PB交于点D,PD=3, 设∠ACB=θ,则∠APM=2θ,又∠ACB=θ,∴C在圆上． ∴AD•DC=BD•DM=BD•（PM+PD）=1•（4+3）=7, 所以答案为7,
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