题目内容
如图,在△ABC中,AB=3,AC=5,点M是BC的中点,AD是∠BAC的平分线,MF∥AD,则FC的长为______.
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如图,设点N是AC的中点,连接MN,则
MN∥AB,MN=
AB.
∴∠CNM=∠BAC.
∵MF∥AD,
∴∠DAC=∠MFN.
∵AD是∠BAC的平分线,∠CNM=∠MFN+∠FMN,
∴∠MFN=∠FMN.
∴FN=MN=
AB,
∴FC=FN+NC=
AB+
AC=4.
故答案为4.
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MN∥AB,MN=
1 |
2 |
∴∠CNM=∠BAC.
∵MF∥AD,
∴∠DAC=∠MFN.
∵AD是∠BAC的平分线,∠CNM=∠MFN+∠FMN,
∴∠MFN=∠FMN.
∴FN=MN=
1 |
2 |
∴FC=FN+NC=
1 |
2 |
1 |
2 |
故答案为4.
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