Answer :
Answer:
1, 4, 5
Step-by-step explanation:
Consider right triangle JKL with right angle JKL and height KM. This height divides triangle into two right triangles KML and KMJ.
Note that
[tex]m\angle MKL=90^{\circ}-m\angle MLK=m\angle KJM\\ \\m\angle JKM=90^{\circ}-m\angle KJM=m\angle MLK[/tex]
So, by AA theorem, we have three similar triangles
[tex]\triangle JKL\sim \triangle JMK\sim \triangle KML[/tex]
So, options 1, 4 and 5 are true
