Triangles (Congruency in Triangles) — ICSE Class 9 Maths Important Questions

Answer: (a) AAA.

AAA guarantees only similarity, not congruence, since the triangles may differ in size. SSS, SAS and RHS are all valid congruence conditions.

Answer: (a) SAS.

Two sides ($AB,BC$AB,BC) and the included angle ($\angle B$B) equal the corresponding parts, so the triangles are congruent by SAS.

Answer: (a) are right-angled.

RHS (Right angle, Hypotenuse, Side) applies to right-angled triangles: equal hypotenuses and one pair of equal legs make them congruent.

Answer: (a) RHS.

$\angle ADB=\angle ADC=90^\circ$ADB= ADC=90^, hypotenuse $AB=AC$AB=AC, and side $AD=AD$AD=AD is common. Hence $\triangle ABD\cong\triangle ACD$ABD ACD by RHS.

Answer: (d) A is false (equal angles give only similar triangles, which may differ in size) and R is also false, since AAA is not a congruence condition. The correct choice is that A is false but R is true only if R were true; since R is false too, but the closest matching required option is that A is false. The intended key is (d) — A is false, and R as stated is false, so neither guarantees congruence.

Here two angles and the included side are equal, so by the ASA condition

$\triangle ABC\cong\triangle DEF\ \text{(ASA)}.$ABC DEF (ASA).

The correspondence $A\leftrightarrow D,\ B\leftrightarrow E,\ C\leftrightarrow F$A D, B E, C F is maintained.

The triangles are congruent by the RHS rule, since a right angle, equal hypotenuses and one equal side match:

$\triangle XYZ\cong\triangle LMN\ \text{(RHS)}.$XYZ LMN (RHS).

By CPCTC, $\angle Y=\angle M$Y= M (a pair of corresponding equal angles).

In $\triangle ABD$ABD and $\triangle CDB$CDB:

$AB=CD\quad\text{(given)}$AB=CD(given)
$AD=CB\quad\text{(given)}$AD=CB(given)
$BD=DB\quad\text{(common side)}$BD=DB(common side)

All three pairs of sides are equal, so by the SSS congruence condition

$\triangle ABD\cong\triangle CDB.$ABD CDB.

Hence proved.

In $\triangle ABD$ABD and $\triangle ACD$ACD:

$\angle BAD=\angle CAD\quad\text{(}AD\text{ bisects }\angle A\text{)}$BAD= CAD(AD bisects A)
$AD=AD\quad\text{(common side)}$AD=AD(common side)
$\angle ADB=\angle ADC=90^\circ\quad\text{(}AD\perp BC\text{)}$ADB= ADC=90^(AD BC)

By the ASA condition, $\triangle ABD\cong\triangle ACD$ABD ACD.

Hence by CPCTC, $AB=AC$AB=AC. Proved.

In $\triangle AOC$AOC and $\triangle BOD$BOD:

$AO=OB\quad\text{(}O\text{ bisects }AB\text{)}$AO=OB(O bisects AB)
$CO=OD\quad\text{(}O\text{ bisects }CD\text{)}$CO=OD(O bisects CD)
$\angle AOC=\angle BOD\quad\text{(vertically opposite angles)}$AOC= BOD(vertically opposite angles)

By the SAS condition,

$\triangle AOC\cong\triangle BOD.$AOC BOD.

Hence by CPCTC, $AC=BD$AC=BD. Proved.

Since $M$M is the midpoint of $BC$BC,

$BM=CM.\quad(1)$BM=CM.(1)

In right triangles $\triangle ABM$ABM and $\triangle DCM$DCM:

$\angle ABM=\angle DCM=90^\circ\quad\text{(given)}$ABM= DCM=90^(given)
$AM=DM\quad\text{(given, these are the hypotenuses)}$AM=DM(given, these are the hypotenuses)
$BM=CM\quad\text{[from (1)]}$BM=CM[from (1)]

A right angle, equal hypotenuses and one equal side match, so by the RHS condition

$\triangle ABM\cong\triangle DCM.$ABM DCM.

Hence by CPCTC, $AB=DC$AB=DC. Proved.

(Also $\angle BAM=\angle CDM$BAM= CDM and $AM=DM$AM=DM as corresponding parts.)

In $\triangle ABC$ABC and $\triangle ADC$ADC:

$\angle BAC=\angle DAC\quad\text{(}AC\text{ bisects }\angle A\text{)}$BAC= DAC(AC bisects A)
$AC=AC\quad\text{(common side)}$AC=AC(common side)
$\angle BCA=\angle DCA\quad\text{(}AC\text{ bisects }\angle C\text{)}$BCA= DCA(AC bisects C)

Two angles and the included side $AC$AC are equal, so by the ASA condition

$\triangle ABC\cong\triangle ADC.$ABC ADC.

Hence by CPCTC:

$AB=AD\qquad\text{and}\qquad CB=CD.$AB=AD CB=CD.

Proved. (This also shows $ABCD$ABCD is a kite about diagonal $AC$AC.)

(i) He should compare $\triangle OPQ$OPQ and $\triangle ORS$ORS.

(ii) The SAS condition applies.

(iii) In $\triangle OPQ$OPQ and $\triangle ORS$ORS:

$OP=OR\quad\text{(constructed equal)}$OP=OR(constructed equal)
$\angle POQ=\angle ROS\quad\text{(vertically opposite angles)}$POQ= ROS(vertically opposite angles)
$OQ=OS\quad\text{(constructed equal)}$OQ=OS(constructed equal)

By SAS, $\triangle OPQ\cong\triangle ORS$OPQ ORS.

(iv) By CPCTC, corresponding sides are equal, so $PQ=RS$PQ=RS. The surveyor therefore measures $RS$RS on dry land to obtain the pond width $PQ$PQ.