Questions: Triangle Congruence: HL (Hypotenuse-Leg)

Question 1

Right triangles ABC and DEF have right angles at C and F. AB = DE = 10 and AC = DF = 6. Which statement is justified?

• A. The triangles are congruent by SAS
• B. The triangles are congruent by HL
• C. No conclusion is possible because this is SSA
• D. The triangles are similar but not necessarily congruent

Answer: (B)

AB and DE are the hypotenuses (they are opposite the right angles), and AC and DF are legs. Hypotenuse and leg congruent in two right triangles gives congruence by HL. The given angle is not between the two sides, so SAS does not apply.

Question 2

Because SSA is not a valid congruence test, the Hypotenuse-Leg pattern also fails to prove congruence.

• T. True
• F. False

Answer: False

HL is a case of SSA that always works. With a right angle, the Pythagorean theorem gives the third side from the other two, so both triangles have all three sides equal and are congruent by SSS.

Question 3

A right triangle has hypotenuse 13 and one leg 5. Explain why every right triangle with these measurements is congruent to it.

Model answer: By the Pythagorean theorem the other leg is √(13² - 5²) = √144 = 12, so every such triangle has sides 5, 12, 13 and they are congruent by SSS.

The right angle removes the ambiguity that SSA normally has. Only one value of the missing leg is possible, so the triangle is determined.

Question 4

From a point P outside a circle with center O, tangent segments PA and PB touch the circle at A and B. Which reasoning shows PA = PB?

• A. Triangles OAP and OBP are congruent by HL, since angles OAP and OBP are right, OP is a shared hypotenuse, and OA = OB
• B. Triangles OAP and OBP are congruent by AAA
• C. Triangles OAP and OBP are congruent by SSA with no further facts needed
• D. PA = PB because all segments from P to the circle are equal

Answer: (A)

A tangent is perpendicular to the radius at the point of tangency, so both triangles are right. They share hypotenuse OP and have congruent legs OA = OB (radii). By HL the triangles are congruent, and corresponding parts give PA = PB.