Triangles — Exercise 6.3

Use only given lengths and angle markings, not the apparent shape of the drawings.

(i)

Angles 60°,80°,40° correspond in order: A↔P,B↔Q,C↔R. AAA (or AA) proves similarity.

(ii)

AB/QR=2/4=1/2; BC/RP=2.5/5=1/2; AC/QP=3/6=1/2. SSS gives A↔Q,B↔R,C↔P.

(iii)

Sorted side ratios are 2/4=1/2, 2.7/5=0.54 and 3/6=1/2. They are unequal, so SSS similarity fails and the triangles are not similar.

(iv)

MN/QP=2.5/5=1/2 and ML/QR=5/10=1/2; their included angles M and Q are both 70°. SAS gives MNL∼QPR.

(v)

The equal 80° angles are not included between the two given side pairs. The displayed data supplies no AA, SSS or SAS correspondence; the non-included-angle pattern does not justify similarity.

(vi)

Triangle DEF has angles D=70°,E=80°,F=30°. Triangle PQR has P=70°,Q=80°,R=30°. AAA gives DEF∼PQR.

OD and OB are opposite rays, so ∠DOC=180°−125°=55°. In triangle DOC, ∠DCO=180°−70°−55°=55°. From ODC∼OBA, C corresponds to A; hence ∠OAB=∠DCO.

Since AB∥CD, ∠OAB=∠OCD and ∠OBA=∠ODC. Hence triangles OAB and OCD are similar by AA. Their corresponding sides give OA/OC=OB/OD (=AB/CD).

In triangle PQR, the marked equal angles PQR=PRQ imply PQ=PR. Substitute PR=PQ in the given relation: QR/QS=QT/PQ, so QS/QR=PQ/QT. The included angle PQS equals TQR because QP and QT are the same ray, and QS and QR are the same ray. SAS proves the correspondence P↔T,Q↔Q,S↔R.

The given equality is ∠RPQ=∠RTS. Also ∠PRQ=∠TRS since RS lies on RP and RT on RQ. Therefore two angle pairs agree. AA gives R↔R,P↔T,Q↔S.

From the congruence in the specified order, AB=AC and AE=AD. Thus AD/AB=AE/AC. The included angle DAE equals BAC because D lies on AB and E lies on AC. SAS similarity applies.

Use right angles from the altitudes, vertically opposite angles at P, and the shared angle at B.

(i) AEP∼CDP

∠AEP=∠CDP=90°; ∠APE=∠CPD are vertically opposite. AA applies.

(ii) ABD∼CBE

∠ADB=∠CEB=90°; ∠ABD=∠CBE because BD lies on BC and BE lies on BA. AA applies.

(iii) AEP∼ADB

∠AEP=∠ADB=90°. Also ∠PAE=∠DAB because AP lies on AD and AE on AB. AA applies.

(iv) PDC∼BEC

∠PDC=∠BEC=90°; ∠PCD=∠BCE because CP lies on CE and CD on CB. AA applies.

AB∥CD, so AB∥CF. Also AE∥BC, since AE lies on AD. Finally BE and BF lie on the same line. Thus ∠BAE=∠FCB and ∠ABE=∠CFB. AA gives A↔C,B↔F,E↔B.

∠ABC=∠AMP=90°. The angles at A agree because AB and AP are the same ray, while AC and AM are the same ray. Therefore ABC∼AMP by AA, with C↔P. The hypotenuses CA,PA correspond, as do legs BC,MP.

The original correspondence is A↔F,B↔E,C↔G. Therefore ∠ACB=∠FGE, and their halves ACD and FGH are equal. Also ∠CAD=∠GFH, since D,H lie on AB,FE. Thus DCA∼HGF by AA; it gives CD/GH=AC/FG. On the other side, ∠DCB=∠HGE (halves) and ∠DBC=∠HEG (original corresponding angles), so DCB∼HGE by AA.

AD⊥BC gives ∠ADB=90°, and EF⊥AC gives ∠EFC=90°. Since AB=AC, the base angles ABC and BCA are equal. With BD along BC, CE along CB, and CF along CA, these equalities give ∠ABD=∠ECF. Thus AA proves A↔E,B↔C,D↔F.

Write AB/PQ=BC/QR=AD/PM=k. Since D,M are midpoints, BD/QM=(BC/2)/(QR/2)=k. Hence ABD∼PQM by SSS. Therefore ∠ABD=∠PQM, which is the same as ∠ABC=∠PQR because BD and QM lie along their base sides. Now AB/PQ=BC/QR and the included B,Q angles are equal. SAS gives the whole-triangle similarity.

Triangles ADC and BAC have ∠ADC=∠BAC by hypothesis. At C, ∠ACD=∠BCA since CD lies on CB. Thus ADC∼BAC by AA, with A↔B,D↔A,C↔C. Corresponding side ratios give AC/BC=DC/AC. Multiply by AC×BC.

Let AB/PQ=AC/PR=AD/PM=k. Extend AD to E so DE=AD, and PM to N so MN=PM. Since D bisects both AE and BC, ABEC is a parallelogram; similarly PQNR is a parallelogram. Therefore BE=AC and QN=PR. Also AE=2AD and PN=2PM, so AB/PQ=BE/QN=AE/PN=k. SSS gives ABE∼PQN, hence ∠BAE=∠QPN. As AE lies along AD and PN along PM, ∠BAD=∠QPM. Together with AB/PQ=AD/PM, SAS gives ABD∼PQM and BD/QM=k. Thus BC/QR=2BD/(2QM)=k. All three side ratios now agree, so ABC∼PQR by SSS.

The vertical objects and level shadows form right triangles. Simultaneous sunlight gives the same angle of elevation, so the triangles are similar by AA. Their height/shadow ratios agree: h/28=6/4.

Original similarity gives AB/PQ=BC/QR and ∠ABC=∠PQR. Since D,M are midpoints, BD/QM=BC/QR. Thus AB/PQ=BD/QM with included angles ABD=PQM. SAS gives ABD∼PQM. Corresponding sides now yield AD/PM=AB/PQ.