The ratio of $\triangle ABH : \triangle ADH$

Question

$AB = AC = 2AD, H$ intersection of $BD$ and $AE$. $AE$ is orthogonal to $BD$. I get that $BH : HD = 2:1$ so the ratio of $\triangle ABH : \triangle ADH$ is $2:1$ but the answer is $4:1$ where am I wrong?

Edit : how I get $BH:HD$ Let $B(0,1), A(0,0), C(1,0)$ $AH$ is a line with equation $y=1/2x$ while the equation of line $BD$ is $y=-2x+1.$ Thus, $BH:HD=2:1$

While the are of triangle is $1/2.$base*height. Both $\triangle ABH : \triangle ADH$ have the same height $AH$. And their base ratio is $2:1$

Answer 1

Let $AB=a$, thus $AD=\frac a2$.

$\angle BAD=90^o\to BD=\sqrt{AB^2+AD^2}=\frac{a\sqrt5}2$

$[ABD]=\frac{AB\cdot AD}2=\frac{AH\cdot BD}2\to AH=\frac{AB\cdot AD}{BD}=\frac{a\sqrt5}5$

By phytagorean theorem in $\Delta ABH$ and $\Delta ADH$, we get $BH=\frac{2a\sqrt5}{5}$ and $HD=\frac{a\sqrt5}{10}$.

$$\frac{[ABH]}{[ADH]}=\frac{BH}{HD}=4:1$$

I think you have made a mistake somewhere in calculating the length ratio.

Answer 2

It may be easier to solve it geometrically. Note that the right triangles ABH and ADH are similar, which leads to the ratios,

$$\frac {HD}{AH}=\frac {AH}{HB} = \frac {AD}{AB}=\frac12\implies \frac{HD}{HB}=\frac{\frac12 AH}{2AH}=\frac14$$

You did not compute the ratio correctly. Note that the lines $y=\frac12 x$ and $y=-2x+1$ intersect at $H(\frac25,\frac15)$. Therefore, with $B(1,0)$ and $D(\frac12,0)$,

$$\frac{HD}{HB}=\frac{\sqrt{(\frac25-\frac12)^2+(\frac15)^2}}{\sqrt{(\frac25)^2+(\frac15-1)^2}}=\frac14$$