In ∆abc, the altitudes from vertex b and c intersect at point m, so that bm = cm. prove that ∆abc is isosceles.

BCM is isosceles. Due to the fact that BM=CM

An altitude in the third vertex A of the triangle ABC, has to go through M, and it will intersect BC in the middle.

This means that the angles <CBA = <BCA, implies that ABC is isosceles.

step-by-step explanation:

the first 5 right here the next 5 i will upload as separate question

what is the original number? you forgot to add it.