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.

Let each base angle be $x$x. The two base angles are equal and

$40^\circ+x+x=180^\circ\Rightarrow 2x=140^\circ\Rightarrow x=70^\circ.$40^+x+x=180^ 2x=140^ x=70^.

Each base angle is $70^\circ$70^.

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, $\angle C=\angle B=(2x+10)^\circ$C= B=(2x+10)^.

Angle sum:

$3x+(2x+10)+(2x+10)=180.$3x+(2x+10)+(2x+10)=180.

$7x+20=180\Rightarrow 7x=160\Rightarrow x=\frac{160}{7}\approx22.86.$7x+20=180 7x=160 x=160/722.86.

Then $\angle A=3x\approx68.6^\circ$A=3x68.6^ and $\angle B=\angle C=2x+10\approx55.7^\circ$B= C=2x+1055.7^.

(Check: $68.6+55.7+55.7\approx180^\circ$68.6+55.7+55.7180^.)

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.

Given: $\triangle ABC$ABC with $AB=AC$AB=AC.

To prove: $\angle B=\angle C$B= C.

Construction: Draw $AD$AD, the bisector of $\angle A$A, meeting $BC$BC at $D$D.

Proof: 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)
$AD=AD\quad\text{(common side)}$AD=AD(common side)

By the SAS congruence condition,

$\triangle ABD\cong\triangle ACD.$ABD ACD.

Hence by CPCTC, $\angle ABD=\angle ACD$ABD= ACD, that is

$\angle B=\angle C.$B= C.

Proved. (The base angles opposite the equal sides are equal.)

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^.)