July | 2017 | Red Manifold

In $\Delta ABC, P, Q$ and $R$ are the midpoints of $AC, AB$ and $BC$ respectively. Lines $PR$ and $QS$ intersect at $T$ . If $BS\perp AC$ , prove that $TS=TP$ .

Solution

Connect points Q and R. Then by midpoint theorem, $AC\parallel QR$ and $2QR=AC$ .

Let M be the intersection of SB and QR.

Since SB is the height of $\Delta ABC$ , then MB is the height of $\Delta QBR$ (i.e. $QR\perp MB$ ). It follows that M is the midpoint of SB and $\Delta SRM\cong \Delta BRM$ by SAS postulate. Hence, $SR=BR$ . As R is the midpoint of CB, then $SR=CR$ , and so $\angle RCS\cong\angle RSC$ . Also, $\angle RSP=180-\angle RSC=180-\angle RCS$ .

Due to the midpoint theorem, PQRC is a parallelogram, which implies that $\angle RCS\cong\angle PQR$ .

In trapezoid PQRS, opposite angles $\angle RSP$ and $\angle PQR$ add up to 180. Therefore PQRS is a cyclic quadrilateral.

Now $\angle PSQ\cong\angle PRQ$ (angle subtended by arc PQ). $SP\parallel RQ$ , then $\angle PRQ\cong\angle SPR$ . Then $\angle PSQ\cong\angle SPR$ . Therefore, $TS=TP$ .

Addendum:
Clearly, a cyclic trapezoid would be formed also by connecting the midpoint of each side and the perpendicular foot on another side.

Theorem: The quadrilateral formed by the midpoint of each side and a perpendicular foot of a triangle is a cyclic trapezoid.

In $\Delta XYZ$ , the lengths of two altitudes are $a$ and $b$ , where $a>b$ . Show that the third side altitude is of range $\left(\frac{ab}{a+b},\frac{ab}{a-b}\right)$ .

Solution:
Let the sides of $\Delta XYZ$ be $s_1, s_2, s_3$ , and the corresponding altitudes be $a,b,c$ , respectively.

Assuming $a>b$ implies that $s_2 > s_1$ .

We then have $as_1=bs_2=cs_3 \Rightarrow c=\frac{as_1}{s_3}=\frac{bs_2}{s_3}$ .

By triangle inequality, $s_2-s_1<s_3<s_2+s_1$ , and thus $\frac{1}{s_2+s_1}<\frac{1}{s_3}<\frac{1}{s_2-s_1}$ .

Writing $\frac{1}{s_2+s_1}<\frac{1}{s_3}$ in terms of the altitudes, we obtain

$\frac{bs_2}{s_2+s_1}<\frac{bs_2}{s_3} \Rightarrow \frac{b}{1+\frac{s_1}{s_2}}<c \Rightarrow \frac{b}{1+\frac{b}{a}} <c \Rightarrow \frac{ab}{a+b}<c$ .