Question 2: State giving reasons, whether the following pairs of triangles are congruent or Not.

$i) \ \ \Delta ABC \ in \ which \angle A = 50^{\circ}, \ \angle B=60^{\circ} \ and \ BC = 4.5 \ cm \ and \ \Delta DEF \\ \ in \ which \ \angle E = 60^{\circ}, \ EF = 4.5 cm, \ \angle F = 70^{\circ}$

$\angle C = 70^{\circ}, \angle D^{\circ} = 50$

$\angle B = \angle E,\ \angle C = \angle F$

$BC = EF = 4.5$

Congruent by ASA

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$ii)\ \ \Delta DEF \ in\ which\ \angle E=48^{\circ},\ DE=6cm\ and\ EF=8cm\ and \ \Delta MNR \ in \ \\ which\ \angle R=48^{\circ},\ MN = 6 cm\ and\ MR = 8cm$

$DE = MN,\ EF = MR \ but \angle E \ne \angle M \ Hence \ Not \ Congruent.$

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$iii)\ \ \Delta KLM \ in \ which \ KM=4 cm, \ \angle K=75^{\circ}, \ \angle M=40^{\circ} \ and\ \Delta PQR \ in \ \\ which \ PR=4 cm,\ \angle Q=65^{\circ},\ \angle R=40^{\circ}$

$\angle L = 65^{\circ} and \angle P = 75^{\circ} \angle K = \angle P,$

$\angle M = \angle R \ and \ KM = PR$

ASA Theorem applied, Triangles are Congruent.

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$iv) \\ \Delta ABC\ in\ which\ AB=3 cm,\ \angle A=90^{\circ},\ BC=5 cm\ and\ \Delta KLM \ in\ \\ which\ KM = 3 cm,\ \angle K=90^{\circ},\ \angle M=5cm$

Applying Pythagoras theorem,

$AC = 4 cm \ KL = 4 cm$

$AB=KM,\ AC=KL\ and\ \angle A=\angle K$

Hence Triangles are congruent by SAS

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Question 3: In the adjoining figure, P in the mid point of AB and $\angle PAC = \angle PBD$. Prove that: $\Delta PAC \cong \Delta PBD$

$\angle PAC = \angle PBD$, (given)

$AP = PB$  (given)

$\angle APC = \angle DPB$ (opposite angles)

Hence $\Delta PAC \cong \Delta ADC$

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Question 4: In the adjoining figure, $DC \parallel AB \ and \ \angle B = \angle D$. Prove that $\Delta ABC \cong \Delta ADC$

Given $\angle B = \angle D$

$\angle DCA = \angle CAB (alternate \ angles)$

$\angle DAC = \angle ACB (alternate \ angles)$

$AC = Common$

Therefore, by ASA $\Delta ABC \cong \Delta ADC$

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Question 5: In the adjoining figure, $BC = AD, \angle CAB = \angle ABD \ and\ \angle ACB = \angle BDA.$

$Prove \Delta ABC \cong \Delta BAD$

In $\Delta ABD \ and\ \Delta ABC$

$\angle CAB = \angle DBA$

$\angle ADB = \angle ACB$

and DA = CB

Therefore, by ASA, $\Delta ABC \cong \Delta BAD$

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Question 6: In the adjoining figure, ABC is a triangle in which $AB=AC,\ BL \perp AC \ and \ CM \perp AB. \ Prove\ that\ BL=CM$.

Take $\Delta ABL \ and\ \Delta ACM$

$AB = AC$

$\angle A$ is common and

$\angle ABL = \angle ACM$

Therefore, by ASA, $\Delta ABL \cong \Delta ACM \ Hence \ BL = CM$

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Question 7: $In \ \Delta ABC AB=AC. \ If \ D \ in \ the \ mid-point \ of \ BC,$

$Prove \ that: \ AD is \ bisector \ of \angle A \ and \ AD \perp BC$

$In\ \Delta ABD \ and\ \Delta ADC$

$AB=AC$

$BD = DC$

$and AD \ is \ Common$

By SSS, $\Delta ABD \cong \Delta ADC$

Therefore $\angle BAD = \angle DAC$

$i) AD \ in \ bisector \ of \angle A$

$ii) \angle ADB + \angle ADC = 180^{\circ}\ \rightarrow \angle ADB = \angle ADC = 90^{\circ}$

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Question 8: In $\Delta ABC$, it in given that $AB = AC \ and\ AD$ in bisector of $\angle A$, meeting BC at D. Prove: i) $\Delta ABD \cong \Delta ACD \ ii) AD \perp BC$

$Given \ AB = AC$

$\angle ABD = \angle ACD$ (angles opposite equal sides of a triangle)

$\angle BAD = \angle DAC$ (angle bisector)

Therefore by ASA:  $\Delta ABD \cong \Delta ACD$

Hence $\angle ADB = \angle ADC = 90^{\circ}\ Therefore,\ AD \perp BC$

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Question 9: In the adjoining figure, $\Delta ABC$ in such that $AB = AC and \angle OBC = \angle OCB$. Prove that: $i) \Delta ABO \cong \Delta ACO \ ii) AO \ in \ the \ bisector \ of \ \angle A$

$\angle ABC = \angle ACB \ given \ AB = AC$

$\angle ABO + \angle OBC = \angle ACO+\angle OCB$

Given $\angle OBC = \angle OCB$

Therefore,

$\angle ABO = \angle ACO$

AB = AC (given)

AO is Common

Because  $\angle OBC=\angle OCB \rightarrow OB=OC$

Therefore, SAS applies, Hence $\Delta ABO \cong \Delta ACO$

Since $\Delta ABO \cong \Delta ACO, \ i) \angle BAO = \angle CAO \ ii) AO \ in \ Bisector \ of \ \angle A$

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Question 10: In the adjoining figure, $AB \parallel GF \ and \ AC \parallel DE \ and \ BF = CE.$

$Prove \ that \ \Delta BDE \cong \Delta FGC$

In $\Delta BDE \ and\ \Delta FGC$

$\angle DBE = \angle GFC$  (alternate angles)

$\angle DEB = \angle GCF$  (alternate angles)

$BF + FE = FE + EC$

Hence, $BE = FC$

Given $BF=EC \ and \ FE$  is common.

Therefore, $BE = FC$

Applying ASA, $\Delta BDE \cong \Delta FGC$

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Question 11: In the adjoining figure, ABCD as a square and CEB is an isosceles triangle in which EC = EB show: $\ i) \Delta DCE \cong \Delta ABE \ ii) AE = DE$

Given $DC = AB$ (Sides of a square)

$CE = EB$ (Sides of an isosceles triangle)

$\angle ECB = \angle EBC$

$Add \ 90^{\circ} on \ both \ side, \angle ECB + 90^{\circ} = \angle EBC + 90^{\circ} or \angle DCE = \angle ABE$

Hence SAS applies,

Therefore, $i)\ \Delta DCE \cong \Delta ABE \ ii) \ Because \ of i)\ AE = DE$

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Question 12: Find the vales of x and y in each of the following cases:

i)    Given $PN \parallel LM,$

$PQ = KM$

$\angle PQN = \angle MKL$

$\angle PNL = \angle NLM$(alternate angles)

Therefore, AAS applies Hence, $\Delta PQN \cong \Delta LMK$

Therefore $19=3x+4 \Rightarrow x=5 2y-3=55 \Rightarrow y=29$

ii)    In $\Delta ABC \ \& \ \Delta ADC$

$AC = Common$

$AB = AD \ and \ BC = DC$

Therefore by SSS, $\Delta ABC \cong \Delta ADC$

Hence, $7y + 1 = 180-103-34$

$7y+1=43 \Rightarrow y=6 \ and\ 2x-5=103 \Rightarrow x=54$

iii)   In $\Delta PQS \ \&\ \Delta PQR$

$PR = SQ$

$PS = QR$

And PQ in common

Therefore, by SSS, $\Delta PQS \cong \Delta PQR$

Hence $2y-5=71 \Rightarrow y = 38$

$\Delta SPO \cong \Delta ORQ$

Therefore $\ \angle SPO = \angle RQO = 34 \ \& \ \angle SOP = 75^{\circ} \ \&\ \angle POQ = 105^{\circ}$

Since, $\angle SPQ = \angle RQP \angle OPQ = (3x+3)$

Now Calculate, $105 + 2(3x+3) = 180 \Rightarrow x = 11.5$

iv)   In $\Delta RMP \ and\ \Delta RQN$

$\angle RPQ = \angle RQP$

$\angle RPM = \angle RQN$

$RP = RQ \ \&\ \angle MRP = \angle QRN$

Hence, ASA applies, Therefore, $\Delta RPM \cong \Delta RQN$

Hence $x -3=14 \Rightarrow x = 15 \$

And $5y-3=3x+1 \ or \ 5y = 55 \Rightarrow y=11$

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Question 13: In adjoining figure, the sides BA and CA of  have been produced to D and E such that BA = AD and CA = AE Prove ED ∥ BC

In $\Delta AED \ and\ \Delta ABC$

$\angle EAD = \angle BAC$  (Opposite Angles)

$BA = AD \ and\ EA = AC$

Hence $\Delta AED \cong \Delta ABC$

$\Rightarrow \angle ABC = \angle ADE \ and\ \angle ACB = \angle AED$

Hence $ED \parallel BC$

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Question 14: Equilateral triangle ABC and ACE have been drawn on the sides AB and AC respectively of $\Delta ABC$, as shown in the adjoining figure, prove: i) $\angle DAC = \angle EAB \ ii) \ DC = BE$

In $\Delta DAC \ \&\ \Delta EAB$

$DA = AB$

$AC = AE$

$\angle DAC = 60^{\circ} +\angle BAC =\angle BAE = 60^{\circ} +\angle BAC$

Hence $\Delta DAC \cong \Delta EAB$  Therefore,

$i)\ \angle DAC =\angle BAE$

$ii) \ DC = BE$

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Question 15: In a regular pentagon ABCDE, prove that $\Delta ABD \ \$ is isosceles.

Since ABCD in a regular pentagon, all sides are equal and all internal angles are $108^{\circ}$.

In $\Delta ADE \ and\ \Delta BDC$

$ED = DC \ \&\ EA = CB \ \&\ \angle EA = \angle DCB =108^{\circ}$

Therefore $\Delta AED \cong \Delta DCB$

Hence $DA = DB$  Therefore,$\Delta ABD =$ isosceles triangle

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Question 16: In the adjoining figure, $ABCD$ in a square, and $P,\ Q,\ R,\$ are points on the side $AB,\ BC,\ and\ CD$  respectively such that,$AP = BQ = CR \ and\ \angle PQR = 90^{\circ}.$

$Prove: \ i) \ PB = QC \ ii)\ PQ = QR \ iii) \ \angle QPR = 45^{\circ}$

Given ABCD in a square

$\therefore DC = AB$

$DR + RC = AP + PB$

$RC = AP$(given)

$\therefore DR = PB$

Similarly

$DC =BC$

$DR + RC = CQ + QB$

$RC = QB (Given)$

$\therefore DR = CQ$

Now Consider $\Delta PQB \ \&\ \Delta RCQ$

$\angle PBQ = \angle QCR = 90^{\circ}$ (Square)

$\therefore RQ = PQ$

$RC=QB \ \& \ PB=CQ$

$\therefore applying \ SSS, \Delta PQB \cong \Delta RCQ,$

$PQ = QR \Rightarrow \angle QPR = \angle QRP = 45^{\circ}$

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Question 17: In the adjoining figure, $QK \parallel ML, \ QM \parallel KL \ and \ RL=LP.$

$Prove:\ i)\ \angle MPL=\angle KLR \ ii)\ ML=QK$

Since $ML \parallel QK \angle PLM= \angle LRK$

And Similarly, $KL \parallel QM \angle LKR=\angle PML$

$PL = LR$ Given

Therefore, using AAS, $\Delta PML \cong \Delta KLR$

$i) \therefore \angle KLR= \angle MPL$

$ii) \ Since \ QM \parallel KL, \ ML = QK$

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Question 18: In the adjoining figure ABCD is a square, $EF \parallel AC$, and R is mid point of EF. Prove:  $i) \ AE=CF \ ii) \ DE=DF \ iii)\ DR \ bisect\ \angle EDF$

In $\Delta DER \ \&\ \Delta DFR$,

$\angle EDR = \angle FDR$ (Median would bisect the angle)

$ER = FR \ \& \ DR =$ Common

$\therefore \ \Delta DER \cong \Delta DFR$

$\Rightarrow DE = DF$

In $\Delta DAE \ \& \ \Delta DCF$

$DA = DC$ (Side of a square)

$angle DAE = \angle DCF = 90^{\circ} DE = DF$ (Proved above)

Because $\angle EDR=\angle FDR \Rightarrow \angle ADE = \angle CDF$

Applying SAS, Proves that$\Delta DAE \cong \Delta DCF$

Therefore, $AE = CF$. Hence Proven.

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Question 19: In the adjoining figure, $\angle TPQ=\angle SQP,\ \angle SRP=\angle TRQ \ and \ R$ in the Mid point of PQ Prove: $\ i) \ \Delta PRT \cong \Delta QRS \ ii)\ PT = SQ \ iii)\ \angle PTR = \angle QSR$

In $\Delta PRT \ \&\ \Delta QRS$

$\angle TPR = \angle SQR$  (given)

$PR = QR$   (given)

$\angle TRP = \angle SRQ (\angle SRT \ is\ Common)$

$\therefore \ By \ applying \ ASA \ \ \Delta PRT \cong \Delta QRS$

Since $\Delta PRT \cong \Delta QRS, PT = SQ \ \&\ \angle PTR = \angle QSR$

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Question 20: In adjoining figure, $\angle PON = 90^{\circ} and MO = ON.$

$Prove:\ i) \ PM = PN \ ii)\ \angle OMQ = \angle ONQ$

Consider $\Delta MPO \ \&\ \Delta NPO$

$PO$ is common

$MO = ON$ (given)

$\angle POM = \angle PON$

Hence, $\Delta MPO \cong \Delta NPO$

Therefore,  $i) \ PM = PN$

Now Consider $\Delta MOQ \ \&\ \Delta NOQ$

$MO = ON$

$\angle MOQ = \angle NOQ$

$OQ$ is common

Therefore, $\Delta MOQ \cong \Delta NOQ \ Hence, \ \angle OMQ = \angle ONQ$