Fun Geometry Problem with Solution #105



٨ x = 40°
٨ 1



(1)∠BDC = 100°

(2) ˹ش P Ҿз͹ͧش D ҹ BC ∆BCP∆BCD BP = BD, CP = CD,∠CBP =∠CBD = 30° ∠BCP =∠BCD = 50°
∵ BD = BP Р∠DBP = 60° ∆BDP 繠∆ҹ DP = BD DP = AC

(3) ˹ش Q BD DQ = CD (= CP) ∆CDQ 繠∆˹Ҩ ՠ∠D (= 100°) ʹ ∠CQD (= ∠DCQ) = 40°
ѧࡵҠ∆CDQ∆CDP ¤ѹẺ -- (CD = CD,∠CDQ =∠DCP, DQ = CP) CQ = DP CQ = AC ∆ACQ 繠∆˹Ҩ ՠ∠C ʹ ∠CAQ =∠AQC x = 40° Q.E.D.

٨ 2



(1) CD ͡ѧش P ∠CBP = 50°(∠DBP = 20°) ∠CBP =∠BCP ∆BCP 繠∆˹Ҩ ՠ∠P ʹ BP = CP
͡ҡ ѧҠ∠BDP = 80°
ԨóҠ∆BDP Ҡ∠BPD = 80° ∠BPD =∠BDP ∆BDP 繠∆˹Ҩ ՠ∠B ʹ BP = BD BP = AC

(2) ˹ش Q BC PQ⊥BC PQ ǹ٧ͧ∆BCP BQ (= CQ) = BC/2
͡ҡ ѧҠ∠BPQ (=∠CPQ) = (∠BPC)/2 = 40°

(3) ˹ش R AB CR⊥AB
Ҡ∆BCR 繠∆ҡ ՠ∠R ҡ Р∠B = 30° CR = BC/2
ѧࡵҠ∆ACR∆BPQ ¤ѹẺ -- (∠ARC =∠BQP = 90°, CR = BQ, AC = BP) ∠CAR =∠BPQ x = 40° Q.E.D.




Create Date : 02 չҤ 2558
Last Update : 2 չҤ 2558 0:00:00 .
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