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$ABCD$ 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 $AM=\frac{AD\cdot AE}{DE}=12$

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

2 Answers 2

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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.

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  • $\begingroup$ If I show that △ADM and △FAB are similar why does that imply that AF⊥DE? $\endgroup$
    – Birgitt
    Feb 27 at 17:00
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    $\begingroup$ @Birgitt Because that makes $\angle DMA=\angle ABF$. $\endgroup$
    – Arthur
    Feb 27 at 17:01
  • $\begingroup$ Im struggling how to show equality of the non-right angles. I don't know anything about the angles $\endgroup$
    – Birgitt
    Feb 27 at 19:07
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    $\begingroup$ @Birgitt due to interior alternate angles, $\angle DAM = \angle AFB$. Since $\Delta DAE \cong \Delta ABF$, $\angle ADE = \angle BAF$. $\endgroup$
    – D S
    Feb 27 at 19:28
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Since a transformation approach has already been given, here's a vector approach: $$\begin{align} \vec{DE}\cdot \vec{AF} &= (\vec{DA}+\vec{AE}) \cdot (\vec{AB}+\vec{BF}) \\ &= \vec{DA} \cdot \vec{BF} + \vec{AE}\cdot \vec{AB} \\ &= -|DA|\cdot |FB| + |AE|\cdot{AB}| \\ &= 0 \tag*{$\blacksquare$} \end{align}$$

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