Question

• assume that $ac \leq bc$, which means either $ac < bc$ or $ac = bc$
• $\angle b$ is a right $\angle$ (def. of right $\delta$ )
• $m\angle b = \underline{\quad\quad\quad\quad\quad\quad\quad\quad}$ (def. of right $\angle$)
• possibility 1:
• if $ac < bc$, then $m\angle \underline{\quad\quad\quad\quad\quad\quad\quad\quad} < m\angle \underline{\quad\quad\quad\quad\quad\quad\quad\quad}$ (in $\delta$, larger $\angle$ is opposite longer side.)
• this is a contradiction because $\angle a$ is acute so $\underline{\quad\quad\quad\quad\quad\quad\quad\quad}$
• possibility 2:
• if $ac = bc$, then $\underline{\quad\quad\quad\quad\quad\quad\quad\quad} \angle b = m \underline{\quad\quad\quad\quad\quad\quad\quad\quad} a$ (isosceles $\delta$ thrm.)
• this is a contradiction because $\underline{\quad\quad\quad\quad\quad\quad\quad\quad}$
• therefore, since $ac \leq bc$ is false, $ac > bc$ must be true.

a. $\angle$
b. $m$
c. $m\angle b > m\angle a$
d. $90^\circ$
e. $b$
f. $a$
g. $\angle a$ is acute so $m\angle b > m\angle a$
h. $45^\circ$
i. $c

Explanation:

Step1: Find \( m\angle B \)

A right angle has a measure of \( 90^\circ \) by definition. So \( m\angle B = 90^\circ \) (option d).

Step2: Analyze \( AC < BC \) case

In a triangle, the larger angle is opposite the longer side. If \( AC < BC \), then the angle opposite \( AC \) (which is \( \angle B \)) and the angle opposite \( BC \) (which is \( \angle A \)): so \( m\angle B < m\angle A \)? Wait, no—wait, side \( AC \) is opposite \( \angle B \), side \( BC \) is opposite \( \angle A \). Wait, correction: In \( \triangle ABC \), side opposite \( \angle A \) is \( BC \), side opposite \( \angle B \) is \( AC \), side opposite \( \angle C \) is \( AB \). So if \( AC < BC \), then the angle opposite \( AC \) (\( \angle B \)) has measure less than the angle opposite \( BC \) (\( \angle A \))? Wait, no—larger side → larger opposite angle. So if \( AC < BC \), then \( m\angle B < m\angle A \)? But \( \angle B \) is right (90°), \( \angle A \) is acute (less than 90°), so this would mean \( 90^\circ < m\angle A \), but \( \angle A \) is acute (so \( m\angle A < 90^\circ \)), which is a contradiction. Wait, the first blank for the angles: if \( AC < BC \), then \( m\angle B < m\angle A \)? Wait, no, let's re-express. Wait, side \( AC \) is opposite \( \angle B \), side \( BC \) is opposite \( \angle A \). So if \( AC < BC \), then \( m\angle B < m\angle A \) (since larger side → larger opposite angle). But \( \angle B \) is 90°, \( \angle A \) is acute (so \( m\angle A < 90^\circ \)), so \( 90^\circ < m\angle A \) is false, hence contradiction. So the first angle blank: \( \angle B \) (e) and \( \angle A \) (f)? Wait, no—wait, the first dropdown: "if \( AC < BC \), then \( m\angle \underline{\quad} < m\angle \underline{\quad} \)". So side \( AC \) is opposite \( \angle B \), side \( BC \) is opposite \( \angle A \). So if \( AC < BC \), then \( m\angle B < m\angle A \). But \( \angle B \) is 90°, \( \angle A \) is acute, so \( 90^\circ < m\angle A \) is impossible (since \( \angle A \) is acute, \( m\angle A < 90^\circ \)). So the contradiction is because \( \angle A \) is acute, so \( m\angle B > m\angle A \) (option c), which would mean \( 90^\circ > m\angle A \), but if \( AC < BC \), we get \( m\angle B < m\angle A \), which contradicts \( m\angle B > m\angle A \).

Step3: Analyze \( AC = BC \) case

If \( AC = BC \), then \( \triangle ABC \) is isosceles with \( AC = BC \), so the angles opposite them are equal. The angle opposite \( AC \) is \( \angle B \), the angle opposite \( BC \) is \( \angle A \). So \( m\angle B = m\angle A \) (by Isosceles Triangle Theorem: if two sides are equal, their opposite angles are equal). But \( \angle B \) is 90°, so \( m\angle A = 90^\circ \), but \( \angle A \) is acute (so \( m\angle A < 90^\circ \)), which is a contradiction. So the first blank in Possibility 2: " \( m\angle B = m\angle A \)" (so the first dropdown is " \( m\angle \)" (option b) and " \( \angle A \)"? Wait, the options: let's check the options. The first part of Possibility 2: "If \( AC = BC \), then \( \underline{\quad} \angle B = m \underline{\quad} A \)". So using Isosceles Triangle Theorem, if \( AC = BC \), then \( m\angle B = m\angle A \). So the first blank: " \( m\)" (option b), second blank: " \( \angle \)" (option a)? Wait, the options: a. \( \angle \), b. \( m \), c. \( m\angle B > m\angle A \), d. \( 90^\circ \), e. \( B \), f. \( A \), g. \( \angle A \) is acute so \( m\angle B > m\angle A \), h. \( 45^\circ \), i. \( C \). Wait, re-reading: "If \( AC = BC \), then \( \underline{\quad} \angle B = m \underline{\quad} A \) (Isosceles \( \Delta \) Thrm.)". So the first blank is " \( m\)" (option b) (to make \( m\angle B \)), the second blank is " \( \angle \)" (option a) (to make \( m\angle A \)). Then the contradiction: \( \angle A \) is acute, so \( m\angle A < 90^\circ \), but \( m\angle B = 90^\circ \), so \( m\angle B = m\angle A \) would mean \( 90^\circ = m\angle A \), which contradicts \( \angle A \) being acute (so \( m\angle A < 90^\circ \)). So the contradiction is because \( m\angle A \) would be \( 90^\circ \), but \( \angle A \) is acute (so \( m\angle A < 90^\circ \)).

Step4: Fill in the blanks

1. \( m\angle B = \boldsymbol{90^\circ} \) (option d)
2. If \( AC < BC \), then \( m\angle \boldsymbol{B} < m\angle \boldsymbol{A} \) (options e and f)
3. Contradiction because \( \angle A \) is acute so \( \boldsymbol{m\angle B > m\angle A} \) (option c)
4. If \( AC = BC \), then \( \boldsymbol{m}\angle B = m\boldsymbol{\angle} A \) (options b and a)
5. Contradiction because \( \boldsymbol{m\angle A} \) would be \( 90^\circ \) (but \( \angle A \) is acute, so \( m\angle A < 90^\circ \)) (or more precisely, the contradiction is that \( m\angle A \) would equal \( 90^\circ \), contradicting \( \angle A \) being acute)

Answer:

1. \( m\angle B = \boxed{90^\circ} \) (option d)
2. If \( AC < BC \), then \( m\angle \boxed{B} < m\angle \boxed{A} \) (options e, f)
3. Contradiction: \( \angle A \) is acute so \( \boxed{m\angle B > m\angle A} \) (option c)
4. If \( AC = BC \), then \( \boxed{m}\angle B = m\boxed{\angle} A \) (options b, a)
5. Contradiction: \( m\angle A \) would be \( 90^\circ \) (contradicts \( \angle A \) acute)

(Note: The exact filling depends on the dropdown options, but the key answers are as above with the corresponding options.)