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Super difficult math problem in grade two. A reward is being offered.
1. Connect AM, because M is the midpoint of BC and AB=AC, then AM is perpendicular to BC. Let AM = 1, CD = X, MP = 1/2ad = 1/2 roots (AM 2+MD 2).

BE= radical sign ((BC+CD) 2+de 2) = radical sign ((2+x) 2+x 2) = radical sign (2x 2+4x+4).

So MP:BE= 1/2: radical number 2= 1:2.

2. also let AM= 1, CD=DE=x, BD=x-BC=x-2.

MP: Be =1/2ad: Be =1/2 * radical number (am 2+MD 2): radical number (BD 2+de 2) = 1/2 * radical number (1+(2)).

So the conclusion is established.