Proof: Square Example 2
Let's prove the following theorem:
if ABCD is a square, then m∠BCA = 45
Proof:
Given
1 | ABCD is a square |
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# | Claim | Reason |
---|---|---|
1 | ABCD is a rhombus | if ABCD is a square, then ABCD is a rhombus |
2 | m∠BCA = m∠DCA | if ABCD is a rhombus, then m∠BCA = m∠DCA |
3 | m∠BCA = m∠ACD | if m∠BCA = m∠DCA, then m∠BCA = m∠ACD |
4 | ABCD is a rectangle | if ABCD is a square, then ABCD is a rectangle |
5 | ∠BCD is a right angle | if ABCD is a rectangle, then ∠BCD is a right angle |
6 | m∠BCD = 90 | if ∠BCD is a right angle, then m∠BCD = 90 |
7 | quadrilateral ABCD is convex | if ABCD is a rectangle, then quadrilateral ABCD is convex |
8 | point A lies in interior of ∠BCD | if quadrilateral ABCD is convex, then point A lies in interior of ∠BCD |
9 | (m∠BCA) + (m∠ACD) = m∠BCD | if point A lies in interior of ∠BCD, then (m∠BCA) + (m∠ACD) = m∠BCD |
10 | (m∠BCA) + (m∠BCA) = m∠BCD | if (m∠BCA) + (m∠ACD) = m∠BCD and m∠BCA = m∠ACD, then (m∠BCA) + (m∠BCA) = m∠BCD |
11 | (m∠BCA) + (m∠BCA) = 90 | if (m∠BCA) + (m∠BCA) = m∠BCD and m∠BCD = 90, then (m∠BCA) + (m∠BCA) = 90 |
12 | m∠BCA = 45 | if (m∠BCA) + (m∠BCA) = 90, then m∠BCA = 45 |
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