24.如圖,△ABC的兩條高AD與BE交于點(diǎn)O,AD=BD,AC=6.
(1)求證:OB=AC;
(2)若∠ABO=25°,則∠EOD=
°,∠ACB=
°;
(3)點(diǎn)F是射線BC上一點(diǎn),且CF=AO,動(dòng)點(diǎn)P從點(diǎn)O出發(fā),沿線段OB以每秒1個(gè)單位長(zhǎng)度的速度向終點(diǎn)B運(yùn)動(dòng),同時(shí)動(dòng)點(diǎn)Q從點(diǎn)A出發(fā),沿射線AC以每秒4個(gè)單位長(zhǎng)度的速度運(yùn)動(dòng),當(dāng)點(diǎn)P到達(dá)點(diǎn)B時(shí),P,Q兩點(diǎn)同時(shí)停止運(yùn)動(dòng),設(shè)運(yùn)動(dòng)時(shí)間為t秒,當(dāng)△AOP與△FCQ全等時(shí),直接寫(xiě)出t的值.