Can i just say that
"Since the internal angles are equal among the 2 triangles, then the triangles are deemed similar. If so...
Area of BHA = BA = 2AM = 2
Area of AHM AM AM
Since 2AM = BA
Is this a justified answer?
Let Triangle ABC be a right triangle such that <A = 90*, AB = AC and let M be the midpoint of the side AC. Take the point P on the side BC so that AP is vertical to BM. Let H be the intersection point of AP and BM.
Find the ratio between the 2 triangles ABH and AHM
Find the ratio of BP : PC
So far the -BH^2 + HM^2 = 3 is the only equation I was able to formulate :/ Help please.
Thank you! I did not see that!
I FINALLY SOLVED THE FIRST PART. YES! It involved a lot of finding the relationships between the sides of different triangles to one another. My approach was to find the relationship of Triangle BAM with AHM to get AH. Then from there related Triangle BAH to AHM then. Put them into A=bxh/2 and came out 4:1. Is that the easiest/most practical way to solve it? Seems long.
Moving on to part II. I can't find any parameter/relation of point P to anything...
Well, as soon as you prove that triangles ABH and AHM are similar, you say that side AM corresponds to side AH and AM:AH = 1:2
The area is then simply the square, which is 1:4 for triangle ABH:AHM.
2. Oh well, I used some trigonometry for that (sine rule), and it gets quite long, but I get 1:2.25 = 4 : 9 (for PC : AP)
EDIT: To answer this, 3rad5 is