Math

GeometryQ&A Library*2. Prove that if one chord of a circle is bisected by another, either segment of the first is a mean proportional between the two segments of the other. AutoSave Exercises 4_15 proofs 1 and. . Marc Skwarczynski ON Exercise 4.15 #3: A triangle ABC is inscribed in a circle. M is the midpoint of AMČ and BM intersects AC at D. File Home Insert Draw Design Layout References Mailings Review View Help Table Design Layout 1D Exercises 4.15 #3 Given: Inscribed AABC; M midpoint of AMC; BM intersects AC in D AB AD Prove: %3D BC DC AB Prove: BC AD Statements Reasons DC 1. (see above) 1. Given 2. CM = AM 2. Def. of midpoint 3. ZCBM = - CM; 3. An inscribed angle = ½ the intercepted arc. %3D %3D ZABM = AM M. 4. - CM = - AM 4. Division (of step 2) %3D 2. 5. Substitution 5. ZCBM = LABM %3D (step 3 -> step 4) 6. Def. of bisector 6. BM bisects ZABC 7. The bisector of an interior angle of a divides the oppos internally into se which have the sa AB 7. BC AD %3D DC as the other two 257 words fecusStart your trial now! First week only $4.99!*arrow_forward*

Question

Please answer in a two-column proof-statements and reasons, not paragraph.

**Also, it should have ONLY one given (statement).**

I have an example of how it should look like/how it should be done in the second picture.

It is about lines proportional. The question is in the picture. *The question is completely clear, so please don't reject the question.*

*Maybe, if you could, choose some of these reasons to prove the statements true. (Those below are the ideas/reasons for to prove the statements if you want to use)*

Definition of ~ triangles

Definition of bisector

Definition of perimeter

Definition of median

Definition of midpoint

Multiplication

Division

If 4 quantities are in proportion, then like powers are in proportion.

Subtraction Transformation

Alternation Transformation

Pythagorean Theorem

If 2 angles have their sides perpendicular, right side to right side and left side to left side, the angles are equal.

In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent.

Angles inscribed in the same segment or equal segments are equal.

If a line is drawn parallel to the base of a triangle, it cuts off a triangle similar to the given triangle.

Two isosceles triangles are similar if any angle of one equals the corresponding angle of the other.

C.A.S.T.E. - corresponding angles of similar triangles are equal

C.S.S.T.P. - corresponding sides of similar triangles are proportional

Theorem 53-If a line is drawn through two sides of a triangle parallel to the third side, then it divided the two sides proportionally.

Theorem 54-If a line divides two sides of a triangle proportionally, then it is parallel to the third side (Converse of Theorem 53)

Theorem 55-The bisector of an interior angle of a triangle divides the opposite side internally into segments which have the same ratio as the other two sides.

Theorem 56-The bisector of an exterior angle of a triangle divided the opposite side externally into segments which have the same ratio as the other two sides.

Theorem 57- If two triangles have the three angles of one equal respectively to the three angles of the other, then the triangles are similar

Corollary 57-1 If two angles of one triangle are equal respectively to two angles of another, then the triangles are similar. (a.a.)

Corollary 57-2 Two right triangles are similar if an acute angle of one is equal to an acute angle of the other.

Theorem 58-If two triangles have two pairs of sides proportional and the included angles equal respectively, then the two triangles are similar. (s.a.s.)

Corollary 58-1 If the legs of one right triangle are proportional to the legs of another, the triangles are similar. (l.l.)

Theorem 59- If two triangles have their sides respectively proportional, then the triangles are similar. (s.s.s.)

Theorem 60- If two parallels are cut by three or more transversals passing through a common point, then the corresponding segments of the parallels are proportional.

Theorem 61-If in a right triangle the perpendicular is drawn from the vertex of the right angle to the hypotenuse

Corollary 61-1 If a perpendicular is dropped from any point on a circle upon a diameter, then the perpendicular is the mean proportional between the segments of the diameter.

Theorem 62-The square of the hypotenuse of a right triangle is equal to the sum of the squares of the legs.

Corollary 62-1 The difference of the square of the hypotenuse and the square of one leg equals the square of the other leg.

Theorem 63- If two chords intersect within a circle, then the product of the segments of one is equal to the product of the segments of the other.

Corollary 63-1 The product of the segments of any chord through a fixed point within a circle is constant.

Theorem 64-If from a point outside a circle, a tangent and a secant are drawn to the circle, then the tangent is the mean proportional between the secant and its external segment.

Corollary 64-1 The product of any secant from a fixed point outside the circle and its external segment is constant.

Corollary 64-2 If two or more secants are drawn to a circle from a fixed point outside the circle, then the product of one secant and its external segment is equal to the product of any other secant and its external segment.

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Transcribed Image Text

*2. Prove that if one chord of a circle is bisected by another, either segment of the first is a mean proportional between the two segments of the other.

AutoSave Exercises 4_15 proofs 1 and. . Marc Skwarczynski ON Exercise 4.15 #3: A triangle ABC is inscribed in a circle. M is the midpoint of AMČ and BM intersects AC at D. File Home Insert Draw Design Layout References Mailings Review View Help Table Design Layout 1D Exercises 4.15 #3 Given: Inscribed AABC; M midpoint of AMC; BM intersects AC in D AB AD Prove: %3D BC DC AB Prove: BC AD Statements Reasons DC 1. (see above) 1. Given 2. CM = AM 2. Def. of midpoint 3. ZCBM = - CM; 3. An inscribed angle = ½ the intercepted arc. %3D %3D ZABM = AM M. 4. - CM = - AM 4. Division (of step 2) %3D 2. 5. Substitution 5. ZCBM = LABM %3D (step 3 -> step 4) 6. Def. of bisector 6. BM bisects ZABC 7. The bisector of an interior angle of a divides the oppos internally into se which have the sa AB 7. BC AD %3D DC as the other two 257 words fecus