CONGRUENCE OF SIMPLE POLYGONS B6

In the following figure, prove that ∆JKM is congruent to ∆LKM

solution

Given: figure JKLM with K produced to M

Required to prove: ∆JKM ≡ ∆LKM

Proof: KM = KM (common and given)

                  <MKJ = <MKL (given)

                     JK = LK (given)

Hence ∆JKM ≡ ∆LKM by SAS rule (proved)