Exercise: Two Quadrilaterals with Diagonals
Given:
ABCD is a quadrilateral, diagonals AC and BD intersect at O
EFGH is a quadrilateral, diagonals EG and FH intersect at P
AO = CO (O bisects AC)
BO = DO (O bisects BD)
EP = GP (P bisects EG)
FP = HP (P bisects FH)
Each quadrilateral has 4 triangles formed by the diagonals. Let's find congruent triangles!
Part A
Prove that:
△AOB ≅ △COD
By which congruence theorem?
Think: AO=CO (given), BO=DO (given), and the angle between them ∠AOB=∠COD (vertical angles)
Exactly right!
Proof of Part A
△AOB ≅ △COD by SAS
S
AO = CO
(Given - O bisects AC)
A
∠AOB = ∠COD
(Vertical angles)
S
BO = DO
(Given - O bisects BD)
Therefore:
△AOB ≅ △COD
(By SAS congruence theorem)
What can we conclude?
AB = CD
(Corresponding sides of congruent triangles)
Part B - Another Pair of Triangles in Quadrilateral ABCD
Already proven:
△AOB ≅ △COD
Prove that:
△BOC ≅ △DOA
By which congruence theorem?
Again we have Side-Angle-Side! BO=DO, angles ∠BOC=∠DOA are vertical, and CO=AO.
Excellent!
Proof of Part B
△BOC ≅ △DOA by SAS
S
BO = DO
(Given - O bisects BD)
A
∠BOC = ∠DOA
(Vertical angles)
S
CO = AO
(Given - O bisects AC)
Therefore:
△BOC ≅ △DOA
(By SAS congruence theorem)
What can we conclude?
BC = DA
(Corresponding sides of congruent triangles)
Part C - Now in Quadrilateral EFGH
Already proven in Quadrilateral ABCD:
△AOB ≅ △COD
△BOC ≅ △DOA
Prove that:
△EPF ≅ △GPH
By which congruence theorem?
Same principle! EP=GP, angles ∠EPF=∠GPH are vertical, and FP=HP.
Outstanding!
Proof of Part C
△EPF ≅ △GPH by SAS
S
EP = GP
(Given - P bisects EG)
A
∠EPF = ∠GPH
(Vertical angles)
S
FP = HP
(Given - P bisects FH)
Therefore:
△EPF ≅ △GPH
(By SAS congruence theorem)
What can we conclude?
EF = GH
(Corresponding sides of congruent triangles)
Part D - The Fourth Pair in Quadrilateral EFGH
Already proven:
△AOB ≅ △COD, △BOC ≅ △DOA, △EPF ≅ △GPH
Prove that:
△FPG ≅ △HPE
By which congruence theorem?
And again! FP=HP, angles ∠FPG=∠HPE are vertical, and GP=EP.
Great!
Proof of Part D
△FPG ≅ △HPE by SAS
S
FP = HP
(Given - P bisects FH)
A
∠FPG = ∠HPE
(Vertical angles)
S
GP = EP
(Given - P bisects EG)
Therefore:
△FPG ≅ △HPE
(By SAS congruence theorem)
What can we conclude?
FG = HE
(Corresponding sides of congruent triangles)
Well Done!
You completed the entire exercise successfully!
You found all the congruent triangles:
In Quadrilateral ABCD:
△AOB ≅ △COD
△BOC ≅ △DOA
In Quadrilateral EFGH:
△EPF ≅ △GPH
△FPG ≅ △HPE
Total: 8 triangles in 4 congruent pairs!
All congruences were by SAS - thanks to vertical angles!