The Hinge Theorem: Practice

Equality of the first arms requires

Subtract p and subtract 7.

Now the equal pairs surround a smaller angle at A, so

The hinge comparison requires

This gives x greater than 5. Each third side must also be between the difference 5 and the sum 19.

For the first third side,

For the second third side,

Combining all restrictions gives

Every x in this interval yields two valid triangles; their third-side order gives the required angle order by the converse.

The converse hinge theorem requires

Subtract 100, then divide by negative 4, reversing the inequality.

The first angle must stay below 180 degrees.

This gives r greater than negative 20. Its positive-angle condition is weaker than r less than 15.

SAS has made PC equal to PX, so PC also measures 3 units. The stated point order allows segment addition.

Substitute the consecutive lengths.

Therefore,

Ruling out equality leaves two possible strict orders.

If angle A were smaller than angle D, the forward hinge theorem would give

That contradicts the given and excludes the other alternative.

Find the included angle in each triangle from the angle sum.

Angle A measures 75 degrees.

Angle D measures 68 degrees. The hinge theorem therefore gives

The hinge theorem and its converse compare the included angles, not every angle.

If every corresponding angle in one triangle were larger, its angle sum would exceed the other triangle’s 180-degree sum. Both sums must be 180 degrees.

The 120-degree opening exceeds the 60-degree opening, so the hinge theorem gives

The triangle inequality also keeps the third side strictly below the sum of the arms.

Thus c is less than 19.

The hinge theorem needs both corresponding arm pairs to match.

Here AC and DF differ by 2 units. The angle comparison alone does not meet the theorem’s hypotheses.

The converse hinge theorem makes the included angle at A larger than the one at D.

Subtracting each included angle from 180 degrees leaves the corresponding sum of the other two angles. A larger subtracted angle leaves a smaller sum.

Conversely, suppose the sum at B and C is smaller than the sum at E and F. Subtract each sum from 180 degrees. The smaller sum leaves the larger included angle at A. With the two matched side pairs, the forward hinge theorem then gives

This proves both directions of the equivalence.

The hinge theorem gives

The triangle inequality gives the upper bound

The difference bound is already satisfied. Solve the two inequalities.

The possible integers are 8, 9, and 10, so there are 3.

The corresponding third sides are 17, 19, and 21. Each is strictly between the arm difference 5 and arm sum 23, so each forms a triangle with arms 9 and 14. Each also exceeds 16, so the converse hinge theorem gives the required larger included angle. All three values occur.

The two given side pairs and equal included angles establish SAS congruence.

Corresponding third sides therefore have equal lengths.

A strict order would contradict this equality.