## [教學]Special geometric theorems ### [教學]Special geometric theorems

Interior angle bisector ratio theorem ABC is a triangle, D lies on BC, such that ∠BAD=∠CAD, then AB/BD=AC/CD. (Interior angle bisector ratio theorem)

Proof E lies on AB, such AC//ED.  Join DE.

DE//CA (Given)
∴BD/DC=BE/EA (Equal ratios)
∴ED=EA (Side opposite equal angles)
∴BD/DC=BE/ED

∠DBE=∠CBA (Common angle)
∠BED=∠BAC (Corresponding angles, DE//CA)
∠EDB=∠ACB (Corresponding angles, DE//CA)
∴△ABC∼△EBD (AAA)
∴BE/ED=BA/AC (Corresponding sides, similar triangles)
∴BD/DC=BA/AC
∴AB/BD=AC/CD

Converse of Interior angle bisector ratio theorem ABC is a triangle, D lies on BC, such that AB/BD=AC/CD, then ∠BAD=∠CAD. (Converse of Interior angle bisector ratio theorem)

Proof E lies on AB, such AC//ED.  Join DE.

DE//CA (Given)
∴BD/DC=BE/EA (Equal ratios)
AB/BD=AC/CD (Given)
∴BD/DC=AB/AC
∴AB/AC=BE/EA

∠DBE=∠CBA (Common angle)
∠BED=∠BAC (Corresponding angles, DE//CA)
∠EDB=∠ACB (Corresponding angles, DE//CA)
∴△ABC∼△EBD (AAA)
∴BE/ED=BA/AC (Corresponding sides, similar triangles)
∵AB/AC=BE/EA (Proved)
∴EA=ED
∴∠EAD=∠EDA (Base angles of isosceles triangle)
∵∠EDA=∠DAC (Alternate angles, DE//CA)

Exterior angle bisector ratio theorem ABC is a triangle, extend BC and BA to D and E respectively, such that ∠EAD=∠CAD, then AB/BD=AC/CD. (Exterior angle bisector ratio theorem)

Proof F lies on AB, such FC//AD.  Join CF.

∴BF/FA=BC/CD (Equal ratios)
∴AC=AF (Side opposite equal angles)
∴BF/CA=BC/CD
∴BF/BC=AC/CD

∠FBC=∠ABD (Common angle)
∴△BCF∼△BDA (AAA)
∴BF/BC=BA/BD (Corresponding sides, similar triangles)
∴AB/BD=AC/CD

Converse of Exterior angle bisector ratio theorem ABC is a triangle, extend BC and BA to D and E respectively, such that AB/BD=AC/CD, then ∠EAD=∠CAD. (Converse of Exterior angle bisector ratio theorem)

Proof F lies on AB, such FC//AD.  Join CF.

∴BC/CD=BF/FA (Equal ratios)
∴FB/BC=AF/CD
∠FBC=∠ABD (Common angle)
∴△FBC∼△ABD (AAA)
∴FB/BC=AB/BD (Corresponding sides, similar triangles)
∴AB/BD=AF/CD
∵AB/BD=AC/CD (Given)
∴AF=AC
∴∠AFC=∠ACF (Base angles of isosceles triangle)

Ptolemy's Theorem Proof M lies on BD, such that ∠DCM=∠ACB

∠DCM=∠ACB (Given)
∠MDC=∠BAC (Angles in the same segment)
∠CMD=180°-∠DCM-∠MDC (Angle sum of triangle)
∠CBA=180°-∠ACB-∠BAC (Angle sum of triangle)
∴∠CMD=∠CBA
∴△CMD∼△CBA (AAA)
∴DC/DM=AC/AB (Corresponding sides, similar triangles)
∴(DC)(AB)=(DM)(AC) ---(1)

∠DAC=∠MBC (Angles in the same segment)
∠ACD=∠DCM+∠MCE
∠BCM=∠ACB+∠MCE
∴∠ACD=∠BCM
∠CDA=180°-∠DAC-∠ACD (Angle sum of triangle)
∠CMB=180°-∠MBC-∠BCM (Angle sum of triangle)
∴∠CDA=∠CMB
∴△CDA∼△CMB (AAA)
∴MB/BC=DA/AC (Corresponding sides, similar triangles)
∴(MB)(AC)=(BC)(DA) ---(2)

(1)+(2)：
(AC)(DM+MB)=(AB)(CD)+(BC)(DA)

Converse of Ptolemy's Theorem If ABCD is a quadrilateral, and (AB)(CD)+(BC)(AD)=(AC)(BD), then A, B, C, D are concyclic. (Converse of Ptolemy's Theorem)

Proof M lies in ABCD, such that ∠DCM=∠ACB, ∠MDC=∠BAC

∠DCM=∠ACB (Given)
∠MDC=∠BAC (Given)
∠CMD=180°-∠DCM-∠MDC (Angle sum of triangle)
∠CBA=180°-∠ACB-∠BAC (Angle sum of triangle)
∴∠CMD=∠CBA
∴△CMD∼△CBA (AAA)
∴DC/DM=AC/AB (Corresponding sides, similar triangles)
∴(DC)(AB)=(DM)(AC) ---(1)

CM/CB=CD/CA (Corresponding sides, similar triangles)
∠BCM=∠BCE+∠ECM
∠ACD=∠ACM+∠MCD
∴∠BCM=∠ACD
∴△BCM∼△ACD (Ratio of 2 sides, included angle)

(1)+(2)：
∴(AC)(DM+MB)=(AC)(BD)
∴DM+MB=BD
∴BMD is a straight line
∵∠MDC=∠BAC (Given)
∴∠BDC=∠BAC
∴A, B, C, D are concyclic (Converse of Angles in the same segment)

Heron's Formula In triangle ABC, bc=a, CA=b, AB=c.  If s=(1/2)(a+b+c), then the area of △ABC=√[s(s-a)(s-b)(s-c)]。 (Heron's Formula)

Proof H lies on BC, such that AH is perpendicular to BC.  Let AH=h.

AH⊥BC (Given)
∴AB2-AH2=HB2 (Pythagoras Theorem)
∴c2-h2=HB2
∴HB=√(c2-h2)

∴AC2-AH2=HC2 (Pythagoras Theorem)
∴b2-h2=HC2
∴HC=√(b2-h2)

∵BC=BH+HC
∴a=√(c2-h2)+√(b2-h2)
h=±(1/2a)√[2a2(b2+c2)-(b2-c2)2-a4]
∵h＞0
∴h=(1/2a)√[2a2(b2+c2)-(b2-c2)2-a4]

The area of △ABC
=(1/2)(BC)(AH)
=(1/2)ah
=(1/4)√[2a2(b2+c2)-(b2-c2)2-a4]
=(1/4)√[(a+b+c)(b+c-a)(c+a-b)(a+b-c)]
=√(1/16)[(a+b+c)(a+b+c-2a)(a+b+c-2b)(a+b+c-2c)]
=√{[(1/2)(a+b+c)][(1/2)(a+b+c-2a)][(1/2)(a+b+c-2b)][(1/2)(a+b+c-2c)]}
=√{[(1/2)(a+b+c)][(1/2)(a+b+c)-a][(1/2)(a+b+c)-b][(1/2)(a+b+c)-c]}
=√[s(s-a)(s-b)(s-c)]

Common height triangles Definition: Two or more triangles which have a common height

Property of Common height triangles ABC is a triangle, D and E are two points on BC, then Area of △ABD: Area of △ADE: Area of △AEC=BD:DE:EC。 (Property of Common height triangles)

Proof H lies on BC, such that AH is perpendicular to BC

Area of △ABD=(1/2)(BD)(AH)
Area of △AEC=(1/2)(EC)(AH)

∴Area of △ABD: Area of△ADE: Area of △AEC=(1/2)(BD)(AH) : (1/2)(DE)(AH) : (1/2)(EC)(AH)
∵(1/2)(BD)(AH) : (1/2)(DE)(AH) : (1/2)(EC)(AH)=BD:DE:EC
∴Area of △ABD: Area of△ADE: Area of △AEC=BD:DE:EC