If ABC is a right triangle and CD is the altitude upon the hypotenuse, prove:
a) triangle ACD ~ triangle ABC
b) triangle CBD ~ triangle ABC
c) triangle ACD ~ triangle CBD
*draw and label your diagram
statements:
2. CD perpendicular AB
3. angle CDB, angle CDA are right angles
4. triangle ACD and triangle CBD are right triangles
5. <B = <B, <A = <A
6. triangle ACD ~ triangle ABC
triangle CBD ~ triangle ABC
triangle ACD ~ triangle CBD
Source: ABEKA Plane Geometry 2006
2. definition of altitude
3. definition of perpendicular lines
4. definition of a right triangle
5. identity
6. two right triangles with a pair of acute angles = are ~.
Source: ABEKA Plane Geometry 2006
ABCD is a parallelogram and P any point on BC. If AP and DC extended meet at O, prove: triangle ABP ~triangle OCP ~triangle ODA
statements:
2. <O = <O
3. AD parallel to BC
AB parallel to OD
4. <D = <OCP
5. <B = <D
6. <D = <OCP = <B
7. <PAB = <O
8. <O = <O = <PAB
9. triangle ABP ~ triangle OPC ~ triangle OAD
Source: ABEKA Plane Geometry 2006
2. identity
3. definition of parallelogram
4. if 2 parallel lines are cut by transversal, then the corresponding angles formed are equal.
5. opposite angles of a parallelogram are equal.
6. substitution
7. if 2 parallel lines are cut by a transversal, then the alternate interior angles formed are equal.
8. substitution
9. a.a.
Source: ABEKA Plane Geometry 2006
The diagonals of a trapezoid divide each other proportionally.
Given: trapezoid ABCD with diagonal AC, BD intersecting at O
Prove: AO:CO = BO:DO
statements:
2. AB parallel to CD
3. <CDB = <ABD
<DCA = <BAC
4. triangle AOB ~ triangle COD
5. AO:CO = BO:DO
Source: ABEKA Plane Geometry 2006
2. definition of a trapezoid
3. if two parallel lines are but by a transversal, then the alternate interior angles are equal.
4. a.a.
5. C.S.S.T.P.
Source: ABEKA Plane Geometry 2006
From the trisection points of side AB of triangle ABC lines are drawn to C. Prove that these lines trisect all lines parallel to AB and are terminated by the other two sides.
given: triangle ABC
D, E trisection points of AB
CD, CE
RU parallel to AB
prove: CD and CE trisect RU
statements:
2. AD:RS = DE:ST = EB:TU
3. AD = DE = EB
4. AD:RS = AD:ST = AD:TU
5. RS:AD = ST:AD = TU:AD
6. RS = ST - TU
7. CD and CE trisect RU
Source: ABEKA Plane Geometry 2006
2. if 2 parallels are cut by three or more transversals passing through a common point, the corresponding segments of the parallels are proportional
3. definition of the trisection of a segment
4. substitution axiom
5. inversion transformation
6. multiplication axiom (multiply each term by AD which then cancels)
7. definition of trisector
Source: ABEKA Plane Geometry 2006
