The Angle-Bisector Ratio Theorem: Practice

Pair BD with AB and DC with AC.

The whole base contains 14+25 ratio parts; BD uses fourteen of them.

$\begin{aligned}BD&=31.2\cdot\frac{14}{39},\\&=11.2.\end{aligned}$

The opposite-side parts have ratio two to three, so AB is two thirds of AC.

$\begin{aligned}AB&=27\cdot\frac 8{12},\\&=18.\end{aligned}$

Compare the ratios in the same order.

The interior hypothesis allows the converse.

Being longer in the same direction is not enough; the ratios must be equal.

The direct theorem rules out an internal bisector.

Use the fixed endpoint order before cross-multiplication.

The resulting side lengths are $10$, $14$, and $12$, which satisfy the triangle inequality.

DC occupies eleven of the seventeen whole-side ratio parts.

$\begin{aligned}DC&=34\cdot\frac{11}{17},\\&=22.\end{aligned}$

First find the leftover part, then add it to the first part.

The midpoint makes the part ratio 1, so the adjacent sides must be equal.

Equal parallel-line half-angle pairs make triangle AEC isosceles. Thus AE equals AC, which is 16; BE would instead be 9+16=25.

The four-part adjacent side is AB; AC has nine of the same ratio parts.

$\begin{aligned}AC&=18\cdot\frac 94,\\&=40.5.\end{aligned}$

The positive ratio AB/AC is less than 1, so BD/DC is less than 1. The theorem preserves the comparison between the paired lengths.