Isosceles Triangles — ICSE Class 9 Maths Important Questions

Answer: (a) $80^\circ$80°.

Since $AB=AC$AB=AC, $\angle C=\angle B=50^\circ$∠ C=∠ B=50°. Then $\angle A=180^\circ-50^\circ-50^\circ=80^\circ$∠ A=180°-50°-50°=80°.

Answer: (a) $60^\circ$60°.

All three angles are equal and sum to $180^\circ$180°, so each is $\dfrac{180^\circ}{3}=60^\circ$180°/3=60°.

Answer: (a) $PQ=PR$PQ=PR.

By the converse of the isosceles triangle theorem, sides opposite equal angles are equal. Side opposite $\angle Q$∠ Q is $PR$PR and opposite $\angle R$∠ R is $PQ$PQ, so $PQ=PR$PQ=PR.

Answer: (a) $BD>DC$BD>DC.

The bisector of the apex angle of an isosceles triangle bisects the base and is perpendicular to it, so $BD=DC$BD=DC and $AD\perp BC$AD⊥ BC. Hence $BD>DC$BD>DC is false.

Answer: (a) A is a direct case of R (the converse of the isosceles triangle theorem): equal angles $\Rightarrow$⇒ equal opposite sides, so $AB=AC$AB=AC. R correctly explains A.

The greater angle is opposite the greater side. The greatest side is $CA=9$CA=9 cm, so the greatest angle is $\angle B$∠ B. The smallest side is $BC=5$BC=5 cm, so the smallest angle is $\angle A$∠ A.

The interior angle $\angle C=180^\circ-110^\circ=70^\circ$∠ C=180°-110°=70°.

Since $AB=AC$AB=AC, $\angle B=\angle C=70^\circ$∠ B=∠ C=70°.

So $\angle A=180^\circ-70^\circ-70^\circ=40^\circ$∠ A=180°-70°-70°=40°.

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

$AB=AC\quad\text{(given)}$AB=AC (given)
$\angle BAD=\angle CAD\quad\text{(}AD\text{ bisects }\angle A\text{)}$∠ BAD=∠ CAD (AD bisects ∠ A)
$AD=AD\quad\text{(common)}$AD=AD (common)

By SAS, $\triangle ABD\cong\triangle ACD$△ ABD≅ △ ACD, so by CPCTC $\angle ADB=\angle ADC$∠ ADB=∠ ADC.

But $\angle ADB+\angle ADC=180^\circ$∠ ADB+∠ ADC=180° (linear pair on $BC$BC). Hence

$2\angle ADB=180^\circ\Rightarrow \angle ADB=90^\circ.$2∠ ADB=180° ⇒ ∠ ADB=90°.

Proved.

Since $AB=AC$AB=AC, $\angle C=\angle B=(3x+5)^\circ$∠ C=∠ B=(3x+5)°. Angle sum: $4x+2(3x+5)=180\Rightarrow10x+10=180\Rightarrow x=17$4x+2(3x+5)=180⇒10x+10=180⇒ x=17.

$\angle A=68^\circ$∠ A=68° and $\angle B=\angle C=56^\circ$∠ B=∠ C=56°. (Check: $68+56+56=180$68+56+56=180.)

In $\triangle ABQ$△ ABQ and $\triangle ACP$△ ACP:

$AB=AC\quad\text{(given)}$AB=AC (given)
$\angle A=\angle A\quad\text{(common angle)}$∠ A=∠ A (common angle)
$AQ=AP\quad\text{(given)}$AQ=AP (given)

By the SAS condition, $\triangle ABQ\cong\triangle ACP$△ ABQ≅ △ ACP.

Hence by CPCTC, $BQ=CP$BQ=CP. Proved.

(a) No. The sum of any two sides must be greater than the third, but $3+4=7<8$3+4=7<8.

(b) $\angle R=180^\circ-50^\circ-70^\circ=60^\circ$∠ R=180°-50°-70°=60°. The greater side is opposite the greater angle: smallest angle $\angle P$∠ P ($50^\circ$50°) faces $QR$QR, then $\angle R$∠ R ($60^\circ$60°) faces $PQ$PQ, then $\angle Q$∠ Q ($70^\circ$70°) faces $PR$PR. So $QR<PQ<PR$QR<PQ<PR.

(c) No. Of all the straight lines that can be drawn to a given line from a point outside it, the perpendicular is the shortest. Any other path $PN$PN is the hypotenuse of right $\triangle PMN$△ PMN, so it is longer than $PM=12$PM=12 m.

Part 1 ($BD=CE$BD=CE):

Since $D$D is on $BA$BA produced, $BD=BA+AD$BD=BA+AD; similarly $CE=CA+AE$CE=CA+AE.

Given $AB=AC$AB=AC and $AD=AE$AD=AE, adding:

$BA+AD=CA+AE\Rightarrow BD=CE.\quad(1)$BA+AD=CA+AE⇒ BD=CE. (1)

Part 2 (equal angles):

Since $AB=AC$AB=AC, the base angles give $\angle ABC=\angle ACB$∠ ABC=∠ ACB ...(2)

Also, consider $\triangle DBC$△ DBC and $\triangle ECB$△ ECB:

$BD=CE\quad\text{[from (1)]}$BD=CE [from (1)]
$BC=CB\quad\text{(common)}$BC=CB (common)

We also need the included pair. Note $\angle DBC=\angle ABC$∠ DBC=∠ ABC and $\angle ECB=\angle ACB$∠ ECB=∠ ACB are the same base angles measured at $B$B and $C$C (since $D,E$D,E lie on the produced equal sides through $A$A). From (2), $\angle ABC=\angle ACB$∠ ABC=∠ ACB, hence

$\angle DBC=\angle ECB.\quad(3)$∠ DBC=∠ ECB. (3)

Using $BD=CE$BD=CE, $\angle DBC=\angle ECB$∠ DBC=∠ ECB and common $BC$BC, by SAS

$\triangle DBC\cong\triangle ECB,$△ DBC≅ △ ECB,

which confirms $\angle DBC=\angle ECB$∠ DBC=∠ ECB and gives $DC=EB$DC=EB. Proved.

(i) Since $AB=AC$AB=AC, $\angle B=\angle C$∠ B=∠ C. Angle sum:

$\angle B+\angle C+50^\circ=180^\circ\Rightarrow 2\angle B=130^\circ\Rightarrow \angle B=\angle C=65^\circ.$∠ B+∠ C+50°=180° ⇒ 2∠ B=130° ⇒ ∠ B=∠ C=65°.

(ii) In an isosceles triangle the altitude from the apex to the base is also the median and the angle bisector: $\triangle ABD\cong\triangle ACD$△ ABD≅ △ ACD (RHS), so $AD$AD bisects both $\angle A$∠ A and $BC$BC.

(iii) $\angle BAD=\dfrac{1}{2}\angle A=\dfrac{1}{2}(50^\circ)=25^\circ.$∠ BAD=1/2∠ A=1/2(50°)=25°.

(iv) $AD\perp BC$AD⊥ BC, so $\angle ADB=90^\circ$∠ ADB=90°. (Check in $\triangle ABD$△ ABD: $25^\circ+65^\circ+90^\circ=180^\circ$25°+65°+90°=180°.)