Finding the ratio of a dissected isosceles?

Question

I have trouble trying to find relationship between sub-triangles.

Answer 1

For part 1), note that both $\triangle ABH$ and $\triangle AHM$ are right triangles similar to $\triangle ABM$:

$$\triangle AHM \sim \triangle BHA \sim \triangle BAM.$$

Using the formula for the area of a right triangle as one-half the product of its legs, we can calculate directly:

$$\frac{\text{area}(\triangle ABH)}{\text{area}(\triangle AHM)} = \frac{\frac12 (AH)(BH)}{\frac12 (AH)(HM)} = (\frac{BH}{AH})(\frac{AH}{HM})=4.$$

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Hints for part 2): For convenience, choose coordinates for the vertices of the triangle $\triangle ABC$ to be $A=(0,0)$, $B=(1,0)$, and $C=(0,1)$ so that its legs have unit length: $AB=AC=1$. The equation of the line BM is $y=\frac12-\frac12 x$. The line AP is perpendicular to BM, so the slope of AP must be equal to the negative reciprocal of the slope of BM. And since AP goes through the origin, the equation of the line AP is simply $y=2x$. Thus to find the coordinates of the point $P$ where the line $AP$ intersects the line $BC$, we solve the system of equations:

$$\begin{cases}y=2x,\\ y=1-x.\end{cases}$$

Once you've found the coordinates of $P$, it's a simple matter to use the Cartesian distance formula to find the length of either $PC$ or $BP$. Note that since the sum of $PC$ and $BP$ must be $\sqrt{2}$, you don't have to actually compute both lengths directly.