试题:
如图已知△ABC中,∠B和∠C外角平分线相交于点P.
(1)若∠ABC=30°,∠ACB=70°,求∠BPC度数.
(2)若∠ABC=α,∠BPC=β,求∠ACB度数.

答案:

我来补答
(1)∠BPC
=180°-(
1
2
∠EBC+
1
2
∠BCF)
=180°-
1
2
(∠EBC+∠BCF)
=180°-
1
2
(180°-∠ABC+180°-∠ACB)
=180°-
1
2
(180°-30°+180°-70°)
=50°;

(2)∠BPC=180°-
1
2
(180°-∠ABC+180°-∠ACB)
=
1
2
(∠ABC+∠ACB),
∵∠BPC=β,∠ABC=α,
∴β=
1
2
(α+∠ACB).
故∠ACB=2β-α.
 
 
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