Answer:
Step-by-step explanation:
In ΔACB and ΔCAD,
AD║BC and AC is a transverse [Given]
∠BCA ≅ ∠DAC [Alternate interior angles]
AC ≅ AC [Reflexive property]
BC ≅ AD [Given]
ΔACB ≅ ΔCAD [By SAS property of congruence]
From the second figure,
In ΔJHG and ΔLKH,
JH ≅ LK [Given]
JG ≅ LH [Given]
GH ≅ HK [Given]
Therefore, ΔJHG ≅ ΔLKH [By SSS property of congruence]
