Slide3Question 1: In the adjoining figure, X and Y are two points on equal sides AB and AC of \triangle ABC such that AX = AY . Prove that AC = YB .


Consider \triangle AYB and \triangle AXC

AX = AY (given)

\angle A is common

AB = AC (given)

Therefore \triangle AYB \cong \triangle AXC (by S.A.S theorem)

Hence XC = YB


Slide4Question 2: If D is the mid point of the hypotenuse AC of a right triangle ABC , prove that BD =\frac{1}{2} AC .


Construct ABD such that ABD = DE

Consider \triangle ABD and \triangle ECD

AD = DC (given)

\angle ADB = \angle EDC (vertically opposite angles)

BD = DE (constructed)

Therefore \triangle ABD \cong \triangle EDC (by S.A.S theorem)

\Rightarrow EC = AB and \angle ABD = \angle CED

\Rightarrow CE \parallel AB

\Rightarrow \angle ABC + \angle ECB = 180^o (interior angles on the same side of the transversal are supplementary)

\Rightarrow 90^o + \angle ECB = 180 \Rightarrow \angle ECB = 90^o

Therefore \triangle ABC and \triangle ECB are right angle triangles.

Therefore \angle ABC = \angle ECB

BC is common


Therefore \triangle ABC \cong \triangle ECB

\Rightarrow AC = BE

\Rightarrow \frac{1}{2} AC = \frac{1}{2} BE = BD . Hence proved.


Slide5Question 3: In a quadrilateral ACBD, \ AC = AD and AB bisects  \angle A . Show that \triangle ABC \cong \triangle ABD . What can you say about BC and BD .


Consider \triangle ABC   and \triangle ABD

AC = AD (given)

\angle CAB = \angle DAB (given)

AB is common

Therefore \triangle ABC \cong \triangle ABD (by S.A.S theorem)

\Rightarrow BC = BD


Slide6Question 4: Prove that \triangle ABC is isosceles if any of the following holds: (i) Altitude AD bisects BC (ii) Median AD is perpendicular to the base BC .


(i)  If BD = DC

Consider \triangle ADB and \triangle ADC

AD is common

\angle ADB = \angle ADC = 90^o (altitude is perpendicular to the base)

BD = DC (given)

Therefore \triangle ADB \cong \triangle ADC

Therefore AB = AC \Rightarrow \triangle ABC is isosceles triangle.

(ii) If Median AD is perpendicular to the base BC

\angle ADB = \angle ADC = 90^o

AD is common

BD = DC (since AD is the median, D is the midpoint of BC )

Therefore \triangle ADB \cong \triangle ADC 

Therefore AB = AC \Rightarrow \triangle ABC is isosceles triangle.


2018-07-13_8-10-40.jpgQuestion 5: In the adjoining figure PS = QR  and \angle SPQ = \angle RQP  . Prove that \triangle PQS \cong \triangle QPR  , PR = QS  and \angle QPR = \angle PQS  .


Consider \triangle PQS and \triangle PQR

PS=QR is given

\angle SPQ = \angle RQP given

PQ (is common)

Therefore \triangle PQS \cong \triangle PQR

Therefore PR = QS and \angle QPR = \angle PQS


Slide7Question 6: In the adjoining figure, AC = AE , AB = AD and \angle BAD = \angle EAC . Prove that BC = DE


Given \angle BAD = \angle EAC AE = AC and AB = AD

\angle BAD + \angle DAC = \angle DAC + \angle CAE

Therefore \triangle ABC \cong \triangle ADE (by S.A.S theorem)


Slide9.JPGQuestion 7: In a \triangle PQR, if PQ = QR , and L, M and N are midpoints of the sides PQ, QR and RP respectively. Prove that LN = MN .


Given PQ = QR

\Rightarrow PL = LQ = QM = MR


Consider \triangle PLN and \triangle MNR



\angle P = \angle R

\therefore \triangle PLN \cong \triangle MNR

Therefore LN = LM


Slide8.JPGQuestion 8: In the adjoining figure, PQRS is a square and SRT is an equilateral triangle. Prove that (i) PT=QT (ii) \angle TQR = 15^o


Consider \triangle PST and \triangle QRT

ST = RT (equilateral triangle)

SP = RQ (sides of a square)

Since \angle PST = 60^o+90^o = 150^o and \angle QRT = 60^o+90^o = 150^o

\angle PST = \angle QRT

Therefore \triangle PST \cong \triangle QRT

In \triangle TQR


Therefore \angle QTR = \angle TQR = x

\Rightarrow 2x+ 150 = 180 \Rightarrow x = 15^o


2018-07-13_20-42-58Question 9: Prove that the median of an equilateral triangle are equal. 


Given: AD = DB, BF = FC, AE = EC

\triangle ABC is equilateral

Therefore \angle A = \angle B = \angle C = 60^o

To prove: AF = BE = CD

Consider \triangle ABE and \triangle DBC

AB = BC ( \triangle ABC is an equilateral triangle)

\angle A = \angle B ( \triangle ABC is an equilateral triangle)

AE = DB (medians bisect the opposite side. Since \triangle ABC is an equilateral triangle the sides are equal)

Therefore \triangle ABE \cong \triangle DBC

Now consider \triangle ADC and \triangle BEC

\angle A = \angle C ( \triangle ABC is an equilateral triangle)


AC=BC  ( \triangle ABC is an equilateral triangle)

Therefore \triangle ADC \cong \triangle BEC

Hence DC = BE

Hence we can say that AD = EC = BE


Slide11Question 10: AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B . Show that the line PQ is perpendicular bisector of AB .


Consider \triangle APC and \triangle BPC

AP = BP (given)

\angle APC = \angle BPC (bisector)

PC is common

Therefore \triangle APC \cong \triangle BPC

Therefore \angle PCA = \angle PCB = x

\Rightarrow 2x = 180 \Rightarrow x = 90^o

Similarly in \triangle ACQ and \triangle BCQ


\angle AQC = \angle BQC

CQ is common

Therefore \triangle ACQ \cong \triangle BCQ

\Rightarrow \angle ACQ = \angle BCQ = y

\Rightarrow 2y = 180 \Rightarrow 90^o

Therefore PC is a perpendicular bisector of AB


Slide12Question 11: In the adjoining figure, AB = AC and DB = DC , find the ratio \angle ABD  : \angle ACD


Since AB = AC \Rightarrow \angle ABC = \angle ACB

Also since BD = CD \Rightarrow \angle DBC = \angle DCB

Now \angle ABC = \angle ACB

\Rightarrow \angle ABD + \angle DBC = \angle ACD + \angle DCB

\Rightarrow \angle ABD = \angle ADC


2018-07-14_19-22-31Question 12:  In the adjoining figure, \angle BCD = \angle ADC and \angle ACB = \angle BDA . Prove that AD = BC and \angle CAD = \angle CBD .


Given:  \angle BCD = \angle ADC

Since \angle ACB = \angle BDA

Therefore \angle BCD + \angle ACB = \angle ADC  + \angle BDA

\Rightarrow \angle ACD = \angle BDC

Therefore in \triangle ACD and \triangle BDC , we have

\angle ADC = \angle BCD

CD is common

\angle ACD = \angle BDC

Therefore \triangle ACD \cong \triangle ABD (A.S.A. theorem)

Hence AD = BC and \angle CAD = \angle CBD


2018-07-14_19-23-21Question 13: In the adjoining figure, AC = BC, \angle DCA = \angle ECB and \angle DBC = \angle EAC . Prove that \triangle DBC \cong \triangle EAC and hence DC = EC and $latex $latex BD = AE .


Given: \angle DCA = \angle ECB

\Rightarrow \angle DCA + \angle ECD = \angle ECB + \angle ECD

\Rightarrow \angle ECA = \angle DCB

Consider \triangle DBC and \triangle EAC

\angle DCB = \angle  EAC (Proved above)

BC = AC (given)

\angle DBC = \angle EAC (given)

Therefore \triangle DBC \cong \triangle EAC (A.S.A theorem)

Therefore DC = EC and BD = AE


2018-07-19_7-12-48.jpgQuestion 14: In the adjoining figure, it is given that CE = ED, \angle AOC = 2 \angle BCD and \angle BOD = 2 \angle ADC . Prove that \triangle CBE \cong \triangle DAE .


Consider \triangle CED , we have

CE = DE \Rightarrow \angle EDC = \angle ECD … … … … … (i)

We have \angle AOC = \angle BOD (vertically opposite angles)

\Rightarrow 2 \angle BCD = 2 \angle ADC

\Rightarrow \angle BCD = \angle ADC   … … … … … (ii)

Subtracting (ii) from (i) we get

\angle ECD - \angle BCD = \angle EDC - \angle ADC

\Rightarrow \angle ECB = \angle EDA

Therefore in \triangle CBE and \triangle DAE , we have

\angle CEB = \angle DEA (common angle)

CE = DE (given) and

\angle ECB = \angle EDA

Therefore \triangle CBE \cong \triangle DAE (A.S.A. theorem)


2018-07-14_19-24-48Question 15: BD and  CE are bisectors of \angle ABC and \angle ACB of an isosceles \triangle ABC with AB = AC . Prove that BD = CE .


Given: \angle ABD =  \angle DBC (BD bisects the \angle ABC)

Similarly,  \angle ACE =  \angle ECB (EC bisects the \angle ACB)

Since \angle ABC is an isosceles triangle, AB = AC

Consider \triangle ABD and \triangle ACE

\angle BAC is common

Since AB = AC \Rightarrow  \angle ABC =  \angle ACB

\Rightarrow \frac{1}{2} \angle ABC = \frac{1}{2} \angle ACB

\Rightarrow \angle ABD = \angle ACE

Hence \triangle ABD \cong \triangle ACE (A.S.A theorem)