# How to show perpendicularity in a square

$$ABCD$$ is a square so $$AB=BC=CD=DA=20$$ and $$AE=BF=15$$.

Since $$DAE \sphericalangle =90^0$$ we can use the Pythagorean theorem so $$AD^2+AE^2=DE^2$$ and we get that $$DE=25$$.

We know that $$DAE\sphericalangle=ABF\sphericalangle=90^o$$, $$AD=AB=20$$ and $$AE=BF=15$$ so from (SAS) we get that triangles $$DAE\equiv ABF$$.

How to show that $$AF \perp DE$$, so that I can calculate $$AM=\frac{AD\cdot AE}{DE}=12$$

• Prove $\triangle ABF$ and $\triangle ADE$ are congruent. Feb 27 at 16:41
• With coordinate geometry, you can quickly show that the slope multiply to $-1$. Feb 27 at 19:35

Rotate the entire figure $$90^\circ$$ clockwise about the center of the square. $$D$$ will land on where $$A$$ used to be, and $$E$$ will land on where $$F$$ used to be. Thus, $$DE$$ will necessarily land on where $$AF$$ used to be.
Alternately, you can show that the two triangles $$\triangle ADM$$ and $$\triangle FAB$$ are similar by showing equality of the two non-right angles.
• @Birgitt Because that makes $\angle DMA=\angle ABF$. Feb 27 at 17:01
• @Birgitt due to interior alternate angles, $\angle DAM = \angle AFB$. Since $\Delta DAE \cong \Delta ABF$, $\angle ADE = \angle BAF$.
Since a transformation approach has already been given, here's a vector approach: \begin{align} \vec{DE}\cdot \vec{AF} &= (\vec{DA}+\vec{AE}) \cdot (\vec{AB}+\vec{BF}) \\ &= \vec{DA} \cdot \vec{BF} + \vec{AE}\cdot \vec{AB} \\ &= -|DA|\cdot |FB| + |AE|\cdot{AB}| \\ &= 0 \tag*{\blacksquare} \end{align}