Question

Solve the problem. Segment BK (K∈ AC ) is the angle bisector of ∠B in ΔABC. Point M is chosen on the side BC so that MK ≅ MB. Prove KM ∥ AB.

Answer 1

To prove that KM is parallel to AB, we need to show that the alternate interior angles are congruent. Since BK is the angle bisector of angle B, we have angle KBA = angle KBC. Also, since MK is congruent to MB, we have triangle MKB congruent to triangle MBK (by the Side-Angle-Side congruence). From these congruent triangles, we can conclude that angle KMB = angle MBK.

Step-by-step explanation:

To prove that KM is parallel to AB, we need to show that the alternate interior angles are congruent.

Since BK is the angle bisector of angle B, we have angle KBA = angle KBC.

Also, since MK is congruent to MB, we have triangle MKB congruent to triangle MBK (by the Side-Angle-Side congruence).

From these congruent triangles, we can conclude that angle KMB = angle MBK.

Therefore, the alternate interior angles are congruent, and we can conclude that KM is parallel to AB.

Answer 2

Answer: the prove is mentioned below.

Step-by-step explanation:

Here, Segment BK (K∈ AC ) is the angle bisector of ∠B in ΔABC. Point M is chosen on the side BC so that MK ≅ MB

We have to prove that: KM ∥ AB

Since, MK = BM

Therefore, ∠ MKB= ∠ MBK

But, ∠MBK=∠ABK ( Because it is given that BK is the angle bisector)

Therefore By the converse of Interior Alternatiove angle theorem,

KM ∥ AB ( Because ∠MBK and ∠ABK are the angle on lines AB and KM respectively by the same transversal BK)