【摘要】全等三角形證明經(jīng)典題(含答案)1.已知:AB=4,AC=2,D是BC中點(diǎn),111749AD是整數(shù),求ADADBC解:延長(zhǎng)AD到E,使AD=DE∵D是BC中點(diǎn)∴BD=DC在△ACD和△BDE中AD=DE∠BDE=∠ADCBD=DC∴△ACD≌△BDE∴AC=BE=2∵在△ABE中AB-BE<AE<AB+BE∵AB=4
2025-06-22 23:06
【摘要】....全等三角形證明經(jīng)典題(含答案)1.已知:AB=4,AC=2,D是BC中點(diǎn),111749AD是整數(shù),求ADADBC解:延長(zhǎng)AD到E,使AD=DE∵D是BC中點(diǎn)∴BD=DC在△ACD和△BDE中AD=DE∠BDE=∠ADCB
2025-06-22 23:08
【摘要】1.已知:AB=4,AC=2,D是BC中點(diǎn),AD是整數(shù),求ADADBC解:延長(zhǎng)AD到E,使AD=DE∵D是BC中點(diǎn)∴BD=DC在△ACD和△BDE中AD=DE∠BDE=∠ADCBD=DC∴△ACD≌△BDE∴AC=BE=2∵在△ABE中AB-BE<AE<AB+BE∵AB=4即4-2<2AD<4+21<AD<3∴AD=22.已知:D是AB中點(diǎn),∠A
2025-06-22 22:49
【摘要】第一篇:全等三角形證明經(jīng)典10題((含答案) 全等三角形證明經(jīng)典10題(含答案)如圖,已知:AD是BC上的中線(xiàn),且DF=DE.求證:BE∥CF. ,在ΔABC中,D是邊BC上一點(diǎn),AD平分∠BAC...
2024-11-05 12:09
【摘要】第一篇:全等三角形證明經(jīng)典50題(含答案) 全等三角形證明經(jīng)典50題(含答案) :AB=4,AC=2,D是BC中點(diǎn),AD是整數(shù),求AD B D :D是AB中點(diǎn),∠ACB=90°,求證:CD=...
【摘要】第一篇:全等三角形證明經(jīng)典50題(含答案)_ 全等三角形證明經(jīng)典50題(含答案) :D是AB中點(diǎn),∠ACB=90°,求證:CD= 1AB 2延長(zhǎng)CD與P,使D為CP中點(diǎn)。連接AP,BP ∵D...
【摘要】全等三角形證明經(jīng)典50題(含答案)1.已知:AB=4,AC=2,D是BC中點(diǎn),AD是整數(shù),求ADADBC解:延長(zhǎng)AD到E,使AD=DE∵D是BC中點(diǎn)∴BD=DC在△ACD和△BDE中AD=DE∠BDE=∠ADCBD=DC∴△ACD≌△BDE∴AC=BE=2∵在△ABE中AB-BE<AE<AB+BE∵AB=4即
2025-07-29 08:58
2025-06-22 22:55
2025-06-22 22:58
【摘要】ADBC1:已知:AB=4,AC=2,D是BC中點(diǎn),AD是整數(shù),求AD長(zhǎng)。2:已知:D是AB中點(diǎn),∠ACB=90°,求證:DABC:3:已知:BC=DE,∠B=∠E,∠C=∠D,F(xiàn)是CD中點(diǎn),求證:∠1=∠2ABCDEF21BACDF
【摘要】全等三角形證明證明經(jīng)典50題1.已知:AB=4,AC=2,D是BC中點(diǎn),AD是整數(shù),求ADADBC2.已知:D是AB中點(diǎn),∠ACB=90°,求證:DABC3.已知:BC=DE,∠B=∠E,∠C=∠D,F(xiàn)是CD中點(diǎn),求證:∠1=∠2ABCDEF214.已知:∠1=∠2,CD=DE,EF
2025-06-10 15:37
【摘要】1.已知:AB=4,AC=2,D是BC中點(diǎn),AD是整數(shù),求ADADBC2.已知:D是AB中點(diǎn),∠ACB=90°,求證:DABC3.已知:BC=DE,∠B=∠E,∠C=∠D,F(xiàn)是CD中點(diǎn),求證:∠1=∠2ABCDEF214.已知:∠1=∠2,CD=DE,EF//AB,求
2025-03-27 07:41
【摘要】全等三角形經(jīng)典題目精選1.已知:AB=4,AC=2,D是BC中點(diǎn),AD是整數(shù),求ADADBC2.已知:D是AB中點(diǎn),∠ACB=90°,求證:DABC3.已知:BC=DE,∠B=∠E,∠C=∠D,F(xiàn)是CD中點(diǎn),求證:∠1=∠2ABCDEF214.已知:∠1=∠2,CD=DE,EF//A
【摘要】1、如圖,四邊形ABCD是邊長(zhǎng)為2的正方形,點(diǎn)G是BC延長(zhǎng)線(xiàn)上一點(diǎn),連結(jié)AG,點(diǎn)E、F分別在AG上,連接BE、DF,∠1=∠2,∠3=∠4.(1)證明:△ABE≌△DAF;(2)若∠AGB=30°,求EF的長(zhǎng).【解析】(1)∵四邊形ABCD是正方形,∴AB=AD,在△ABE和△DAF中,,∴△ABE≌△DAF.(2)∵四邊形