<PRE><B>Question 1136025</B>
&lt;br&gt;
I can see the justification for the proportion they show relating areas of triangles: ABH/HBP = AHC/HPC.&lt;br&gt;
I don&#39;t see yet how the use of that leads to the answers to the questions that are asked.&lt;br&gt;
Let J and K be on BC so that AJ and HK are perpendicular to BC.&lt;br&gt;
Then using BC (or parts of BC) for the bases, AJ is an altitude of both BAP and PAC, and HK is an altitude of both BHP and PHC.&lt;br&gt;
Then, since BAP and BHP both have base BP, and since PACand PHC have base PC, the ratio of the areas of BAP to BHP is the same as the ratio of the areas of PAC and PHC.&lt;br&gt;
It then follows that the ratio of the areas of ABH and HBP is the same as the ratio of the areas of AHC and HPC.&lt;br&gt;
While I don&#39;t see the path from there to the answers to the questions that are asked, the problem is easily solved using coordinate geometry.&lt;br&gt;
Here is a sketch, with carefully chosen coordinates for the vertices of the original triangle ABC.&lt;br&gt;
{{{drawing(400,250,-10,80,-10,40,
line(0,0,60,0),line(0,0,30,30),line(30,30,60,0),
line(0,0,45,15),line(30,30,40,0),
locate(0,-2,&quot;B(0,0)&quot;),locate(60,-2,&quot;C(60,0)&quot;),locate(30,34,&quot;A(30,30)&quot;),
locate(48,16,M),locate(40,-2,P),locate(32,16,H)
)}}}&lt;br&gt;
M is the midpoint of AC, so M = (45,15).&lt;br&gt;
Then the equation of M is y = (1/3)x.&lt;br&gt;
Line AP is perpendicular to BM, so its slope is -3; since it passes through A(30,30), the equation of AP is y = -3x+120.&lt;br&gt;
Point P is the x-intercept of AP, so P is (40,0).&lt;br&gt;
Point H is the intersection of BM and AP; solving the pair of equations for those lines finds H is (36,12):&lt;br&gt;
{{{drawing(400,250,-10,80,-10,40,
line(0,0,60,0),line(0,0,30,30),line(30,30,60,0),
line(0,0,45,15),line(30,30,40,0),
locate(0,-2,&quot;B(0,0)&quot;),locate(60,-2,&quot;C(60,0)&quot;),locate(30,34,&quot;A(30,30)&quot;),
locate(48,16,&quot;M(45,15)&quot;),locate(40,-2,&quot;P(40,0)&quot;),locate(24,16,&quot;H(36,12)&quot;)
)}}}&lt;br&gt;
Now the questions are easily answered.&lt;br&gt;
(1) ratio of the areas of triangles ABH and AHM&lt;br&gt;
Both triangles have altitude AH; and BH is 4/5 of BM, so the ratio of BH to HM -- and therefore the ratio of the areas of ABH and AHM -- is 4:1.&lt;br&gt;
(2) ratio BP:PC&lt;br&gt;
Simple -- 40:20 = 2:1</PRE>