香港巴士論壇

各位師兄可唔可以教下我做樓上個題 ~
應該係同 angle in alt. segment / tangent prop. 個 d 有關
謝解答

a) proof ABC and PBD are similar triangle first

b) PB=PD and PB+ PA = AC (isso tri)

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會員	時間	評分	評語
atenulization	2012年4月16日 20:20:05	+1	謝版友解答:)

(a) Joining OD,
OD=OC (radii)
Angle ODC= Angle OCD=x ( base angles, isos. triangle ODC)

Angle AOD=2X (ext. angle, triangle OCD)

Joining PO,
Angle AOP=Angle POD=x (tangent properties)

Since AB=AC,
Therefore angle ABC = angle ACB= x (base angles, isos triangle ABC)

angle PDB= angle POD =x (angle in alt. segment)

Since angle PBD=x=angle PDB
Therefore PB=PD (sides opp. equal angles)

(b) Since AC is the diameter,
AC=AB=PA+PB(given),
PB=PD (proved in (a)),
Therefore, PA+PD is equal to the diameter of the circle.

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會員	時間	評分	評語
atenulization	2012年4月16日 21:24:28	+1	thx:)