worksheet 10.2 deductive geometry iii name:

© John Wiley & Sons Australia, Ltd

WorkSHEET 10.2 Deductive geometry III Name: ___________________________ 1 Prove that ÐABC = ÐACB.

AB = AC (given) ÐBAD = ÐCAD (given) AD is common. \ DBAD DCAD (SAS) Because of congruency, \ÐABC = ÐACB (Corresponding angles in congruent triangles are equal.)

2 In the figure below, prove that ÐBAC = ÐDCA

AB = CD (given) BC = DA (given) AC is common. \ DABC DCDA (SSS) Because of congruency,

ÐBAC = ÐDCA (Corresponding angles in congruent triangles are equal.)

3 Prove that ∆𝐴𝐵𝐶 ≅ ∆𝐶𝐷𝐴

AD = BC (given) AC = AC (common) ÐADC = ÐCBA = ⊥ (right angled triangles) \ DADC DCBA (RHS)

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4 In the figure below, prove that DB bisects AC.

ÐDAE = ÐBCE (right angles and given) ÐDEA = ÐBEC (Vertically opposite angles) DE = BE (given) \DADE DCBE (AAS) Because of congruency;

\AE = CE (Corresponding sides in congruent

triangles are equal.)

\DB bisects AC.

5 Prove that if one angle in a parallelogram is 90°, the quadrilateral is a rectangle.

Arbitrarily set ÐADC = 90° (given) \ÐDAB = 90° (Co-interior angles are supplementary as AB DC.) \ÐABC = 90° (Co-interior angles are supplementary as AD BC.) \ÐBCD = 90° (Co-interior angles are supplementary as AB DC.) \ All angles are 90° and two p[airs of parallel sides. \ ABCD is a rectangle.

6 Use congruence to prove that the main diagonal of a kite bisects the angles it passes through.

QP = QR (given) PS = RS (given) QS is common. \DQPS DQRS (SSS) because of congruency we can say, \ÐPQS = ÐRQS and ÐPSQ = ÐRSQ (Corresponding angles in congruent triangles are equal.) \ The main diagonal of a kite bisects the angles

it passes through.

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7

Prove that MNST is a parallelogram.

Because ÐMTS + ÐTSN = 180°, we can say that they are co-interior angles, therefore we can say; \ 𝑀𝑇 ∥ 𝑁𝑆. \MNST is a parallelogram since both pairs of opposite sides are parallel.

8 Prove that the main diagonal of a rhombus bisects the angles it passes through.

MP = MN (Sides in a rhombus are equal.) PQ = NQ (Sides in a rhombus are equal.) MQ is common. \DMPQ DMNQ (SSS) Because of congruency we can say, \ÐPMQ = ÐNMQ and ÐPQM = ÐNQM (Corresponding angles in congruent triangles are equal.) \MQ bisects the angles it passes through.

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9

What sort of quadrilateral is ABCD? Give reasons.

𝐴𝐷 = 𝐵𝐶(𝑔𝑖𝑣𝑒𝑛) ∠𝐴𝐷𝐵 = ∠𝐴𝐵𝐶(𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒)

𝐷𝐵 = 𝐷𝐵(𝑐𝑜𝑚𝑚𝑜𝑛)

∴ ∆𝐴𝐵𝐷 ≅ ∆𝐶𝐷𝐵(𝑆𝐴𝑆) Now, Because of this Congruency, we can say;

∠𝐴𝐵𝐷 = ∠𝐶𝐷𝐵 (these angles correspond to each other inside congruent triangles) Now we can say;

𝐴𝐵 ∥ 𝐵𝐶(𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒𝑎𝑛𝑔𝑙𝑒𝑠) This quadrilateral has two pairs of parallel sides, so, \ABCD is a parallelogram.


10

HO = OF and ÐHGO = ÐOEF. Prove that EFGH is a parallelogram.

HO = OF (given) ÐHGO = ÐOEF (given) ÐEOF = ÐGOH (Vertically opposite angles are equal.) ÐGHO = ÐEFO (180 degrees in each triangle.) \DEOF DGOH (ASA) Because of congruency 𝐸𝑂 = 𝐺𝑂 (corresponding sides in congruent triangles) Now, HO = OF (given) ÐEOH = ÐGOF (Vertically opposite angles are equal.) 𝐸𝑂 = 𝐺𝑂 (proved above) \DEOH DGOF (SAS) Because of congruency ÐFGO = ÐHEO (corresponding angles in congruent triangles.) \ 𝐸𝐻 ∥ 𝐹𝐺 (alternate angles are equal) Similarly ÐFEO = ÐEGO (given) \ 𝐸𝐹 ∥ 𝐻𝐺 (alternate angles are equal) We have two pairs of parallel sides; \EFGH is a parallelogram.

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11 In a quadrilateral, AB = CD, DO = OB and AC ^ BD. Show that ABCD is a rhombus.

AB = CD (given) ÐAOB = ÐCOD (given 90° and vertically opposite) DO = OB (given) \DAOB DCOD (SAS) Because of congruency \AO = CO (Corresponding sides in congruent triangles are equal.) Now we can say, AB = CD (given) ÐAOD = ÐCOB (given 90°) AO = CO (just proved) \DAOD DCOB (SAS) Similarly \DBOC DBOA (SAS) and \DDOA DBOC (SAS) So, all triangles are congruent, hence by congruency, we can say AB = BC = CD = AD (corresponding sides in congruent triangles) \ABCD is a rhombus.

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worksheet 10.2 deductive geometry iii name:

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