# Geometry, Triangles

In the figure, $BC$ is parallel to $DE$. If area of ∆ $PDE$ is $3/7$ of area of ∆ $ADE$, then what is the ratio of $BC$ and $DE$?

I tried finding ratios of height of ∆ $ABC$, $PDE$ & $BPC$, and trying to figure out some commonality, but it didn't​ work out.

P.s. it is not my homework.

Ratio is 5:2. Not sure how.

• and where are the Points $D$ and $E$ situated? – Dr. Sonnhard Graubner Nov 4 '17 at 18:59
• @Dr.SonnhardGraubner refer to image. D lies on AB and E lies on AC. DE is parallel to BC – Ajax Nov 4 '17 at 19:00
• No proof (yet), but ... Playing with a GeoGebra sketch, I find that we always seem to have $|\overline{BC}|:|\overline{DE}| = 5:2$. – Blue Nov 4 '17 at 20:01
• @Blue answer is 5:2. But what do mean? – Ajax Nov 4 '17 at 20:02
• @Blue well. I have that answer written on a piece of paper. Wasn't fully sure. Now I am. ! – Ajax Nov 4 '17 at 20:18

We may assume $$A=(0,0),\quad B=(1,0),\quad C=(0,1), \quad D=(r,0),\quad E=(0,r)$$ for some $r\in\>]0,1[\>$. Intersecting $EB$ with $C D$ gives $P=\bigl({r\over1+r},{r\over1+r}\bigr)$. $ED$ and $PA$ intersect orthogonally at the midpoint $M=\bigl({r\over2},{r\over2}\bigr)$ of $ED$. The ratio of the two triangle areas in question is therefore given by $${|PM|\over |MA|}={\sqrt{2}\bigl({r\over 1+r}-{r\over2}\bigr)\over\sqrt{2}\,{r\over2}}={1-r\over1+r}\ .$$ Since this ratio has to be ${3\over7}$ it follows that $r={2\over5}$.

Here comes my attempt of a geometric derivation of the sought ratio.

Let $M$ be the midpoint of $\overline{BC}$. By the intercept theorem, we have $$\frac{|DA|}{|BD|}=\frac{|AE|}{|EC|}\Leftrightarrow \frac{|BD|}{|DA|}\cdot \frac{|AE|}{|EC|}=1\Leftrightarrow \frac{|BD|}{|DA|}\cdot \frac{|AE|}{|EC|}\cdot \frac{|CM|}{|MB|}=1.$$ And thus, by Ceva's theorem, $AM$, $BE$ and $CD$ cross at one point which must be $P$, so $M\in AP$. Then define $Q,R\in DE$ so that $AQ\perp DE$ and $PR\perp DE$. Then we have $$\frac{|PR|}{|AQ|}=\frac{|PDE|}{|ADE|}=\frac{3}{7}.$$ Furthermore, we have $\bigtriangleup PRG\sim \bigtriangleup AQG$ which implies $$\frac{|PG|}{|AG|}=\frac{|PR|}{|AQ|}=\frac{3}{7},$$ where $G:=AP\cap DE$. Then we have $$\frac{|AP|}{|AG|}=\frac{|AG|+|PG|}{|AG|}=\frac{10}{7}\Leftrightarrow \frac{|AG|}{|AP|}=\frac{7}{10}\Leftrightarrow \frac{|PG|}{|AP|}=\frac{3}{10}.$$ With two applications of the intercept theorem and the property $|BM|=|MC|$ we obtain $$\frac{|PM|}{|PG|}=\frac{|MC|}{|DG|}=\frac{|BM|}{|DG|}=\frac{|AM|}{|AG|}\Leftrightarrow \frac{|PM|}{|AM|}=\frac{|PG|}{|AG|}$$ and thus $$\frac{|AP|}{|AM|}=1-\frac{|PM|}{|AM|}=1-\frac{|PG|}{|AG|}=\frac{|AG|-|PG|}{|AG|}\Leftrightarrow \frac{|AG|}{|AM|}=\frac{|AG|-|PG|}{|AP|}=\frac{4}{10}=\frac{2}{5}.$$ We then use the intercept theorem to deduce $$\frac{|DG|}{|GE|}=\frac{|BM|}{|MC|}=1\Leftrightarrow |DG|=|GE|.$$ Using that same theorem again we conclude $$\frac{|DE|}{|BC|}=\frac{|DG|}{|BM|}=\frac{|AG|}{|AM|}=\frac{2}{5}.$$ And thus, we have $|BC|:|DE|=5:2$, as desired.

At least in the case $|ADE|>|PDE|$ this proof can be easily generalized to Blue's statement $$|PDE|:|ADE|=p:q\Rightarrow |BC|:|DE|=(p+q):|p-q|.$$

Let $h$ be the altitude of triangles $DBC$ and $EBC$ with respect to base $BC$, and $h'$ be the altitude of $ADE$ with respect to base $DE$. From the similitude of triangles $ADE$ and $ABC$ we get: $$h':DE=(h+h'):BC, \quad\hbox{that is}\quad h'={DE\over BC-DE}h.$$ Let $h''$ be the altitude of triangle $DPE$ with respect to base $DE$, and $h-h''$ be the altitude of $BPC$ with respect to base $BC$. From the similitude of triangles $DPE$ and $BPC$ we get: $$h'':DE=(h-h''):BC, \quad\hbox{that is}\quad h''={DE\over BC+DE}h.$$ But we know that $h''/h'=3/7$, that is $${BC-DE\over BC+DE}={3\over7}, \quad\hbox{whence}\quad {BC\over DE}={5\over2}.$$

Let BE meet CD at P. We also let DE be 1 unit and BC = k units, for some k.

According to the given, we can also let [ADE] = 7h and [PDE] = 3h for some non-zero constant h.

Fact-1) When two triangles have the same altitude, the ratio of their areas is proportional to the ratio of their bases.

Then, [PBD] = [PCE] and $\dfrac {[DBP]}{[DPE]} = \dfrac {k}{1}$.

Fact-2) If two objects are similar, the ratio of their areas is equal to the square of ratios of their corresponding sides.

Noting that $\triangle ADE \sim \triangle ABC$ and $\triangle PDE \sim \triangle PCB$, we have

[PBC] = … = $3hk^2$; and [ABC] = … = $7hk^2$.

∴ [BCED] = [ABC] – [ADE] = $7hk^2 – 7h$

[DPB] = $\dfrac {(7hk^2 – 7h) – 3h – 3hk^2}{2} = 2hk^2 – 5h$

$\dfrac {[DBP]}{[DPE]} = \dfrac {k}{1} = \dfrac {2hk^2 – 5h }{3h}$

After eliminating the “h” , we will get $k = \dfrac {5}{2}$ as the only feasible solution from the resultant quadratic.