NCERT Solutions For Class 9 Math Chapter – 7 Exercise 7.3
Q1. ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC If AD is extended to intersect BC at P, Show that
- ∆ABD ∆ACD
- ∆ABP ∆ACP
- AP bisects ∟A as well as ∟D.
- AP is the perpendicular bisector of BC.
Solution:-
- In triangle ABD and ACD.
AB = AC
BD = CD
AD = DA
∆ABD ∆ACD [SSS Rule]
∟BAD = ∟CAD and ∟ABD = ∟ACD [ CPCT]
- In triangle ABP and ACP
AB = AC
AP = PA
∟BAP = ∟CAP
∆ABP ∆ACP
BP = PC ∟BPA = ∟CPA
- ∟BAP = ∟CAP
AP bisects ∟A.
∟BAD + ∟ABD = ∟CAD = ∟ACD
∟BDP = ∟CDP
So , DP bisects ∟D.
AP bisects ∟A as well as ∟D.
- ∟BPA + ∟CPA = 1800
2∟BPA = 1800
∟BPA = 900
AP bisects BC.
Q2. AD is an altitude of an isosceles triangle ABC in which AB = AC Show that
- AD bisects BC.
- AD bisects ∟
Solution:-
∆ABD ∆ACD [RHS Rule]
- BD = CD
AD bisects BC.
- ∟BAD = ∟CAD
AD bisects ∟BAC ∟A.
Q3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR Show that:
- ∆ABM ∆PQN
- ∆ABC ∆
Solution:-
- M and N are mid – points of BC and QR respectively, As AM and PN are medians.
BM = 1/2BC and QN = 1/2QR
Also , BC = QR
1/2BC = 1/2QR = BM = QN
Consider triangles ABM and PQN,
AB = PQ
AM = PN
BM = QN
∆ABM ∆PQN [SSS Rule]
- From Result (i)
∟ABM = ∟PQN
∟ABC = ∟PQR (iii)
AB = PQ
BC = QR
∟ABC = ∟PQR
∆ABC ∆PQR. [SAS rule]
Q4. BE and CF are two equals altitude of a triangle ABC Using RHS Congruence rule , Prove that the triangle ABC is isosceles.
Solution:-
∆BFC ∆CEB [RHS Rule]
∟FBC = ∟ECB
∟ABC = ∟ACB
AC = AB
[ Sides opposite to equal angels of a triangle are equal.]
∆ABC is an isosceles triangle.
Q5. ABC is an isosceles triangle with AB = AC Draw AP perpendicular BC to show that ∟B = ∟C.
Solution:-
AB = BC
AP = PA
∟APB = ∟APC
∆APB ∆ACP
∟B = ∟C.