Triangle ABC is a right isosceles triangle with the right angle at B. Point H lies on side AB, and point D lies on the ray opposite to ray BC with BH = BD. Prove that DH is perpendicular to AC, and that CH is perpendicular to AD.

Step-by-step solution

1. 1

   Set up coordinates at the right angle — that is what makes both parts routine. Because the right angle is at $B$ and the two legs are equal, placing $B$ at the origin with the legs along the axes costs nothing and turns every hypothesis into a coordinate:

   $B=(0,0),\quad A=(a,0),\quad C=(0,a),\qquad a>0$

   $H$ lies on segment $AB$, so $H=(t,0)$ with $0<t\le a$. $D$ lies on the ray opposite ray $BC$, i.e. on the negative $y$-axis, so $D=(0,-d)$ with $d>0$. The condition $BH=BD$ says $t=d$, hence

   $H=(t,0),\qquad D=(0,-t)$

   Notice the single parameter $t$ now encodes the whole configuration.

2. 2

   Prove (a) with a dot product. Form the two direction vectors:

   $\vec{DH}=H-D=(t-0,\,0-(-t))=(t,\,t),\qquad \vec{AC}=C-A=(0-a,\,a-0)=(-a,\,a)$

   Their dot product is

   $\vec{DH}\cdot\vec{AC}=t(-a)+t(a)=-at+at=0$

   Since neither vector is zero ($t>0$ and $a>0$), a zero dot product means the lines are perpendicular, so $DH\perp AC$.

3. 3

   Read (a) again through angles, to see why it is forced. Triangle $BHD$ has the right angle at $B$ (the axes are perpendicular) and legs $BH=BD$, so it is right isosceles and $\angle BHD=\angle BDH=45^{\circ}$. Triangle $ABC$ is right isosceles too, so $\angle BAC=\angle BCA=45^{\circ}$. Line $DH$ therefore meets line $BC$ at $45^{\circ}$ and line $AC$ meets it at $45^{\circ}$ on the other side, and the two $45^{\circ}$ angles add to $90^{\circ}$. The perpendicularity is really a statement about two isosceles right triangles sharing a perpendicular pair of lines.

4. 4

   Prove (b) with the same coordinates. Now use the other pairing:

   $\vec{CH}=H-C=(t,\,-a),\qquad \vec{AD}=D-A=(-a,\,-t)$

   So

   $\vec{CH}\cdot\vec{AD}=t(-a)+(-a)(-t)=-at+at=0$

   Hence $CH\perp AD$. Equivalently, via slopes, $m_{CH}=\dfrac{0-a}{t-0}=-\dfrac{a}{t}$ and $m_{AD}=\dfrac{-t-0}{0-a}=\dfrac{t}{a}$, and their product is $-\dfrac{a}{t}\cdot\dfrac{t}{a}=-1$, the slope criterion for perpendicular lines. The slope form needs $t\neq 0$, which the dot-product form does not, so the dot product is the cleaner argument.

5. 5

   Note the structural reason and verify numerically. Part (b) is really the orthocentre in disguise: in triangle $ACD$ the line $AB$ is an altitude from $A$ (it is perpendicular to $CD$, the $y$-axis) and $H$ lies on it, while $DH\perp AC$ from part (a) makes $DH$ the altitude from $D$; the two altitudes meet at $H$, so the third altitude $CH$ must also pass through it, giving $CH\perp AD$. A numerical check with $a=3.7$, $t=1.4$ returns $\vec{DH}\cdot\vec{AC}=0$ and $\vec{CH}\cdot\vec{AD}=0$ exactly, and the slope product equals $-1.0$.