Quadrilaterals · Ages 14–15
Proofs with quadrilaterals
Find congruent triangles: alternate angles, a common side, opposite sides of the parallelogram.
How to do it
Method
Find congruent triangles: alternate angles, a common side, opposite sides of the parallelogram
∠EAB = ∠ECD, ∠EBA = ∠EDC, AB = CD: △ABE ≅ △CDE (ASA)
Another way
Work back from the end
To prove AE = CE: look at △ABE and △CDE
In full
A proof about a quadrilateral usually finds a pair of congruent triangles in it. In the parallelogram ABCD with its diagonals crossing at E: ∠EAB = ∠ECD (alt. ∠s, AB // DC), ∠EBA = ∠EDC (alt. ∠s, AB // DC), AB = CD (opp. sides of // gram), so △ABE ≅ △CDE (ASA), and AE = CE (corr. sides, ≅△s). One pair a line with its reason; congruence first, and only then its corresponding sides and angles. Proving a parallelogram the other way: showing ∠BCA = ∠DAC gives AD // BC (alt. ∠s equal). Another way: work back from the end; to prove AE = CE, find two triangles holding AE and CE. You may be asked for a missing reason or to put the steps in order.
Worked examples
Have a go first, then open each one for the answer and the working.
Example 1
ABCD is a parallelogram. One line of a proof is: ∠BAC = ∠DCA ( ? ). Which reason goes in the gap?
- A. alt. ∠s, AB // DC
- B. alt. ∠s, AD // BC
- C. opp. ∠s of // gram
- D. vert. opp. ∠s
Choose the right answer
Show the answer and working
Answer: A. alt. ∠s, AB // DC
AC cuts the parallels AB and DC, and ∠BAC and ∠DCA are on either side of it: alt. ∠s, AB // DC.
Example 2
The parallelogram ABCD's diagonals cross at E. In △ABE and △CDE: ∠EAB = ∠ECD, AB = CD and ∠EBA = ∠EDC. So △ABE ≅ △CDE ( ? ). Which condition goes in the gap?
- A. AAS
- B. SAS
- C. ASA
- D. SSS
Choose the right answer
Show the answer and working
Answer: C. ASA
AB lies between ∠EAB and ∠EBA: two angles and the side between, ASA.
Example 3
In the parallelogram ABCD, the bisector of ∠DAB meets DC at F, with AD = 5 cm and AB = 6 cm. FC =
Show the answer and working
Answer: 1 cm
∠DFA = ∠FAB (alt. ∠s, AB // DC) = ∠DAF, so DF = DA = 5 cm (sides opp. equal ∠s); DC = AB = 6 cm, so FC = 6 − 5 = 1 cm.
Practice questions to try
Try these 5; the answers are at the end. For a whole worksheet, see VIP worksheets below.
1.ABCD is a parallelogram. In △ABC and △CDA: AB = CD ( ? ), … Which reason goes in the gap?
- A. given
- B. corr. sides, ≅△s
- C. opp. sides of // gram
- D. common side
Choose the right answer
Answer:
2.ABCD is a parallelogram. One line of a proof is: ∠BAC = ∠DCA ( ? ). Which reason goes in the gap?
- A. opp. ∠s of // gram
- B. alt. ∠s, AB // DC
- C. vert. opp. ∠s
- D. alt. ∠s, AD // BC
Choose the right answer
Answer:
3.The parallelogram ABCD's diagonals cross at E. In △ABE and △CDE: ∠EAB = ∠ECD, AB = CD and ∠EBA = ∠EDC. So △ABE ≅ △CDE ( ? ). Which condition goes in the gap?
- A. SAS
- B. AAS
- C. ASA
- D. SSS
Choose the right answer
Answer:
4.In the parallelogram ABCD, the bisector of ∠DAB meets DC at F, with AD = 9 cm and AB = 13 cm. FC =
Answer:
5.ABCD is a parallelogram whose diagonals cross at E. Prove AE = CE: put the steps in order.
- A. AB = CD (opp. sides of // gram), so △ABE ≅ △CDE (ASA)
- B. AE = CE (corr. sides, ≅△s)
- C. ∠EAB = ∠ECD, ∠EBA = ∠EDC (alt. ∠s, AB // DC)
Write the letters in the right order
Answer:
All the answers
- 1. C. opp. sides of // gram
- 2. B. alt. ∠s, AB // DC
- 3. C. ASA
- 4. 4 cm
- 5. C → A → B
VIP practice notes and papers
Aimed at tests and exams: questions on this topic, new every time, to print or save as a PDF.
- Practice notes: The key point, worked examples step by step, then questions of the same kind to follow, hints fading
- Practice paper: Set out in parts like a school paper, marked, with answers and full working
Start playingPlay and learn in the secondary world, Star Galaxy: animated methods, monster battles, a review of every mistake