如图,已知AB=AD,BC=CD,AC、BD相交于点O,若AB=5,AC=7,BD=6,求∠BCD的度数.
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如图,已知AB=AD,BC=CD,AC、BD相交于点O,若AB=5,AC=7,BD=6,求∠BCD的度数.
如图,已知AB=AD,BC=CD,AC、BD相交于点O,若AB=5,AC=7,BD=6,求∠BCD的度数.
如图,已知AB=AD,BC=CD,AC、BD相交于点O,若AB=5,AC=7,BD=6,求∠BCD的度数.
Eng：
AC and BD cross at the point O
Solution:∠ BCD = 90 °,the following reasons:
∵ AB = AD,BC = CD.
∴ AC ⊥ BD,AC split BD
∴ in the Rt △ ABO in,from the Pythagorean theorem was,
AO ² = AB ²-BO ²
= AB ² - (½ BD) ²
= 5 ² - (½ 6) ²
= 16
Namely,AO = 4
∴ OC = AC-AO
= 7-4
= 3
∵ OC = OD
∴ △ OCD is the isosceles triangle
∴ ∠ OCD = (180 ° -90 °) ÷ 2
= 45 °
That ∠ BCD = ∠ BCO + ∠ OCD
= 2 ∠ OCD
= 2 × 45 °
= 90 °
Where ½ is half of
² for the square
中文：
AC与BD交于点O
∠BCD=90°,理由如下：
∵AB=AD,BC=CD.
∴AC⊥BD,AC平分BD
∴在Rt△ABO中,由勾股定理得,
AO²=AB²-BO²
=AB²-（½BD)²
=5²-（½6）²
=16
即AO=4
∴OC=AC-AO
=7-4
=3
∵OC=OD
∴△OCD是等腰三角形
∴∠OCD=（180°-90°）÷2
=45°
即∠BCD=∠BCO+∠OCD
=2∠OCD
=2×45°
=90°
其中½为二分之一
²为平方
90°。提示：可以证明BD与AC垂直。
BCD=60°