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given: $overline{jm}$ bisects $angle kjl$, $angle jmkcongangle jml$ pro…

Question

given: $overline{jm}$ bisects $angle kjl$, $angle jmkcongangle jml$ prove: $overline{jk}congoverline{jl}$
statements reasons

  1. $overline{jm}$ bisects $angle kjl$ 1.
  2. $angle jmkcongangle jml$ 2.
  3. $angle kjmcongangle ljm$ 3.
  4. $overline{jm}congoverline{jm}$ 4.
  5. $\triangle jmkcong\triangle jml$ 5.
  6. $overline{jk}congoverline{jl}$ 6.

given: $overline{bc}paralleloverline{ef}$, $d$ is the mid - point of $overline{bf}$ prove: $overline{ed}congoverline{cd}$
statements reasons

  1. $overline{bc}paralleloverline{ef}$ 1.
  2. $d$ is the mid - point of $overline{bf}$ 2.
  3. $overline{bd}congoverline{fd}$ 3.
  4. $angle edfcongangle cdb$ 4.
  5. $angle bcdcongangle efd$ 5.
  6. $\triangle bcdcong\triangle efd$ 6.
  7. $overline{ed}congoverline{cd}$ 7.

Explanation:

First proof

Step1: State given information

Given that $\overline{JM}$ bisects $\angle KJL$ and $\angle JMK\cong\angle JML$.

Step2: Use angle - bisector definition

Since $\overline{JM}$ bisects $\angle KJL$, by the definition of an angle - bisector, $\angle KJM\cong\angle LJM$.

Step3: Apply reflexive property

$\overline{JM}\cong\overline{JM}$ by the reflexive property of congruence.

Step4: Prove triangle congruence

With $\angle JMK\cong\angle JML$, $\overline{JM}\cong\overline{JM}$, and $\angle KJM\cong\angle LJM$, $\triangle JMK\cong\triangle JML$ by the ASA congruence criterion.

Step5: Get the final result

Since $\triangle JMK\cong\triangle JML$, by the property that corresponding parts of congruent triangles are congruent (CPCTC), $\overline{JK}\cong\overline{JL}$.

Second proof

Step1: State given information

Given that $BC\parallel EF$ and $D$ is the mid - point of $\overline{BF}$.

Step2: Use mid - point definition

As $D$ is the mid - point of $\overline{BF}$, by the definition of a mid - point, $\overline{BD}\cong\overline{FD}$.

Step3: Identify vertical angles

$\angle EDF$ and $\angle CDB$ are vertical angles, so $\angle EDF\cong\angle CDB$.

Step4: Use parallel line property

Since $BC\parallel EF$, $\angle BCD\cong\angle EFD$ as alternate interior angles.

Step5: Prove triangle congruence

With $\angle EDF\cong\angle CDB$, $\overline{BD}\cong\overline{FD}$, and $\angle BCD\cong\angle EFD$, $\triangle BCD\cong\triangle EFD$ by the ASA congruence criterion.

Step6: Get the final result

Since $\triangle BCD\cong\triangle EFD$, by CPCTC, $\overline{ED}\cong\overline{CD}$.

Answer:

First proof:

  1. Given
  2. Given
  3. Definition of angle - bisector (If a ray bisects an angle, it divides the angle into two congruent angles)
  4. Reflexive property of congruence (Any segment is congruent to itself)
  5. Angle - Side - Angle (ASA) congruence criterion ($\angle JMK\cong\angle JML$, $JM = JM$, $\angle KJM\cong\angle LJM$)
  6. Corresponding parts of congruent triangles are congruent (CPCTC)

Second proof:

  1. Given
  2. Given
  3. Definition of mid - point (If a point is the mid - point of a segment, it divides the segment into two congruent segments)
  4. Vertical angles are congruent
  5. Alternate interior angles are congruent (Since $BC\parallel EF$, alternate interior angles formed by the transversal are congruent)
  6. Angle - Side - Angle (ASA) congruence criterion ($\angle EDF\cong\angle CDB$, $BD = FD$, $\angle BCD\cong\angle EFD$)
  7. Corresponding parts of congruent triangles are congruent (CPCTC)