Triangle Proportionality and Midsegments: Practice

Use first part over leftover part on both sides.

Use part-to-whole on both sides.

Both part-to-part ratios reduce to four sevenths.

The nondegenerate triangle and interior conditions allow the side-splitter converse.

The midsegment length is half the third side.

The resulting lengths are 14 and 28, both positive.

DE matches BC, EF matches AB, and FD matches AC. Halve each corresponding outer length.

The second-to-first transversal factor is eight fifths for each corresponding gap.

$\begin{aligned}QR&=\frac 85\cdot 9,\\&=14.4.\end{aligned}$

ASA congruence in each strip gives equality of the corresponding intercepted lengths.

The first part uses three steps and the leftover part five, so the whole auxiliary length needs eight. The parallel through the third step transfers the first three parts to AX.

The third of six equal parts leaves three equal parts on the other side.

$\begin{aligned}3:(6-3)&=3:3,\\&=1:1.\end{aligned}$

The whole has thirteen ratio parts, and AX uses four of them.

$\begin{aligned}AX&=\frac 4{13}\cdot 39,\\&=12.\end{aligned}$

The ordinary converse used here requires both points to lie inside their named sides. E beyond C does not meet that hypothesis.

AE is five of the twelve parts of the whole AC.

$\begin{aligned}AC&=15\cdot\frac{12}{5},\\&=36.\end{aligned}$

The leftover EC would be 21; that is not the requested whole length.