Question

given: jm ≅ lk, kj ∥ lm, jm ∥ lk
prove: △jmk ≅ △lkm
given: xv ≅ xz, ∠v ≅ ∠z
prove: ∠w ≅ ∠y
given: ab ≅ ad, ac ⊥ bd
prove: △abc ≅ △adc

Explanation:

First Proof ($\triangle JMK\cong\triangle LKM$)

Step1: Justify $\angle JMK\cong\angle LKM$

Since $JK\parallel LM$ and $JM\parallel LK$, when two parallel lines are cut by a transversal, alternate - interior angles are congruent. So, $\angle JMK\cong\angle LKM$ (alternate - interior angles).

Step2: Justify $MK\cong KM$

By the reflexive property of congruence, any segment is congruent to itself. So, $MK\cong KM$.

Step3: Justify the triangles congruence

We know that $JM\cong LK$ (given), $\angle JMK\cong\angle LKM$ (alternate - interior angles), and $MK\cong KM$ (reflexive property). By the Side - Angle - Side (SAS) congruence criterion, $\triangle JMK\cong\triangle LKM$.

Second Proof ($\angle W\cong\angle Y$)

Step1: Justify the third pair of angles

Since $\angle VXW$ and $\angle ZXY$ are vertical angles, $\angle VXW\cong\angle ZXY$.

Step2: Justify the triangles congruence

We have $XV\cong XZ$ (given), $\angle V\cong\angle Z$ (given), and $\angle VXW\cong\angle ZXY$ (vertical angles). By the Angle - Side - Angle (ASA) congruence criterion, $\triangle VXW\cong\triangle ZXY$.

Step3: Justify $\angle W\cong\angle Y$

If two triangles are congruent, then their corresponding parts are congruent. So, $\angle W\cong\angle Y$ (corresponding parts of congruent triangles).

Answer:

For the first proof:

1. Alternate - interior angles (because \(JK\parallel LM\) and \(JM\parallel LK\) with transversal \(MK\))
2. Reflexive property of congruence
3. \(\triangle JMK\cong\triangle LKM\) (by SAS)

For the second proof:

1. Given
2. \(\angle VXW\cong\angle ZXY\)
3. \(\triangle VXW\cong\triangle ZXY\) (by ASA)
4. Corresponding parts of congruent triangles are congruent

For the third proof (already mostly filled):

1. Given
2. Definition of perpendicular (\(AC\perp BD\))
3. Right angles are congruent
4. \(AC\cong CA\)
5. Hypotenuse - Leg (HL) congruence criterion (since \(AB\cong AD\) (hypotenuse) and \(AC\cong CA\) (leg) and \(\angle ACB\cong\angle ACD = 90^{\circ}\))