Triangles : Exercise 7.3 (Mathematics NCERT Class 9th)


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Q.1     \Delta ABC and \Delta DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that
               (i) \Delta ABD \cong \,\Delta ACD
               (ii) \Delta ABP \cong \,\Delta ACP
               (iii) AP bisects \angle A as well as \angle D
               (iv) AP is the perpendicular bisector of BC.
20Sol.

(i) In \Delta s ABD and ACD, we have
AB = AC [Given]
BD = DC [Given]
and AD = AD [Common]
Therefore by SSS criterion of congruence, we have \Delta \,ABD \cong \Delta ACD

(ii) In \Delta s ABP and ACP , we have
AB = AC [Given]
\angle BAP = \angle PAC
[Since \Delta \,ABD \cong \,\Delta \,ACD
 \Rightarrow \angle BAD = \angle DAC    \Rightarrow   \angle BAP = \angle PAC
and    AP = AP [Common]
Therefore By SAS criterion of congruence , we have  \Delta \,ABP \cong \,\Delta \,ACP

(iii) Since \Delta \,ABD \cong \,\Delta \,ACD. Therefore , \angle BAD = \angle DAC
 \Rightarrow AD bisects \angle A  \Rightarrow AP bisects \angle A ... (1)
In \Delta s BDP and CDP, we have
BD = CD                   [Given]
BP = PC [Since \Delta \,ABP\, \cong \,\Delta ACP\,  \Rightarrow BP = PC]
and DP = DP [Common]
By SSS criterion of congruence, we have
\Delta \,BDP\, \cong \,\Delta CDP\,
 \Rightarrow \angle BDP = \angle PDC
 \Rightarrow DP bisects \angle D  \Rightarrow AP bisects \angle D ... (2)
Combining (1) and (2), we get
AP bisects \angle A as well as \angle D

(iv) Since AP stands on BC
Therefore \angle APB + \angle APC = 180º [Linear Pairs]
But \angle APB = \angle APC [Proved above]
Therefore \angle APB = \angle APC = {{{{180}^o}} \over 2} = {90^o}
Also BP = PC [Proved above]
 \Rightarrow AP is perpendicular bisector of BC.


Q.2     AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
            (i) AD bisects BC               (ii) AD bisects \angle A.

Sol.

AD is the altitude drawn from vertex A of an isosceles \Delta ABC to the opposite base BC so that AB = AC, \angle ADC = \angle ADB = 90º.
Now, in \Delta s\, ADB and ADC, we have
Hyp. AB = Hyp. AC [Given]
AD = AD [Common]
and \angle ADC = \angle ADB [Since Each = 90º]
Therefore By RHS criterion of congruence, we have

21\Delta \,ADB \cong \,\Delta ADC
 \Rightarrow BD = DC and \angle BAD = \angle DAC
[Since corresponding parts of congruent triangle are equal]
Hence, AD bisects BC, which proves (i) and AD bisects \angle A, which proves (ii).


Q.3     Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \Delta PQR (see figure). Show that :
          (i) \Delta ABM \cong \Delta PQN
          (ii) \Delta ABC \cong \Delta PQR

23Sol.

Two \Delta s ABC and PQR in which AB = PQ, BC = QR and AM = PN.
Since AM and PN are median of \Delta s ABC and PQR respectively.
Now, BC = QR [Given]
 \Rightarrow {1 \over 2}BC = {1 \over 2}QR (Median divides opposite sides in two equal parts)
 \Rightarrow BM = QN ... (1)

Now , in \Delta s ABM and PQN we have
AB = PQ [Given]
BM = QN [From (i)]
and AM = PN [Given]
Therefore By SSS criterion of congruence, we have
\Delta \,ABM\, \cong \,\Delta PQN, which proves (i)
 \Rightarrow \angle B = \angle Q ... (2)
[Since, corresponding parts of congruent triangle are equal]
Now, in \Delta s ABC and PQR we have
AB = PQ [Given]
\angle B = \angle Q [From (2)]
BC = QR [Given]
Therefore By SAS criterion of congruence, we have
\Delta \,ABC\, \cong \,\Delta PQR, which proves (ii)


 Q.4      BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Sol.

In \Delta s BCF and CBE, we have
\angle BFC = \angle CEB [Since Each = 90º]
Hyp. BC = Hyp. BC [Common]
FC = EB
12Therefore By R.H.S. criterion of congruence, we have
\Delta \,BCF\, \cong \,\Delta \,CBE
 \Rightarrow \angle FBC\, = \angle ECB
[Since corresponding parts of congruent triangles are equal]
Now, in \Delta ABC
\angle ABC = \angle ACB [Since \angle FBC = \angle ECB]
 \Rightarrow AB = AC
[Since sides opposite to equal angles of a triangle are equal]
Therefore \Delta \,ABC is an isosceles triangle.


Q.5      ABC is an isosceles triangle with AB = AC. Draw AP  \bot BC to show that \angle B = \angle C
Sol.

In \Delta s ABP and ACP, we have
AB = AC [Given]
AP = AP [Common]
and \angle APB = \angle APC [Since each = 90º]
Therefore , by R.H.S. criterion of congruence we have
26\Delta \,ABP \cong \,\Delta \,ACP
 \Rightarrow \angle B = \angle C
[Since corresponding parts of congruent triangles are equal]



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