Triangles
Chapter 7
Exercise 7.3
EX 7.3 QUESTION 1.
ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that
(i) ΔABD ΔACD
(ii) ΔABP ΔACP
(iii) AP bisects A as well as D.
(iv) AP is the perpendicular bisector of BC.
Solution:
(i) In ∆ABD and ∆ACD,
AB = AC [Given]
AD = DA [Common]
BD = CD [Given]
∴ ∆ABD ≅ ∆ACD [By SSS congruency]
∠BAD = ∠CAD [By C.P.C.T.] …(1)
(ii) In ∆ABP and ∆ACP,
AB = AC [Given]
∠BAP = ∠CAP [From (1)]
∴ AP = PA [Common]
∴ ∆ABP ≅ ∆ACP [By SAS congruency]
(iii) Since, ∆ABP ≅ ∆ACP
⇒ ∠BAP = ∠CAP [By C.P.C.T.]
∴ AP is the bisector of ∠A.
Again, in ∆BDP and ∆CDP,
we have BD = CD [Given]
DP = PD [Common]
BP = CP [ ∵ ∆ABP ≅ ∆ACP]
⇒ A BDP = ACDP [By SSS congruency]
∴ ∠BDP = ∠CDP [By C.P.C.T.]
⇒ DP (or AP) is the bisector of ∠BDC
∴ AP is the bisector of ∠A as well as ∠D.
(iv) ∆ABP ≅ ∆ACP
⇒ ∠APS = ∠APC, BP = CP [By C.P.C.T.]
But ∠APB + ∠APC = 180° [Linear Pair]
∴ ∠APB = ∠APC = 90°
⇒ AP ⊥ BC, also BP = CP
∴ AP is the perpendicular bisector of BC.
EX 7.3 QUESTION 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects A.
Solution:
(i) In ∆ABD and ∆ACD
AB =AC [Given]
∠ADB = ∠ADC [Each 90°]
AD = DA [Common]
∴ ∆ABD ≅ ∆ACD [By RHS congruency]
So, BD = CD [By C.P.C.T.]
⇒ D is the mid-point of BC or AD bisects BC.
(ii) ∠BAD = ∠CAD [By C.P.C.T.]
So, AD bisects ∠A
EX 7.3 QUESTION 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 ΔPQR (see Fig. 7.40). Show that:
(i) ΔABM ΔPQN
(ii) ΔABC ΔPQR
Solution:
BC = QR [Given]
½BC = ½QR
⇒ BM = QN [AM and PN are median.]
(i) In ∆ABM and ∆PQN,
AB = PQ , [Given]
AM = PN [Given]
BM = QN [proved above]
∴ ∆ABM ≅ ∆PQN [By SSS congruency]
(ii) In ∆ABM ≅ ∆PQN [proved above]
⇒ ∠B = ∠Q [By C.P.C.T.]
Now, in ∆ABC and ∆PQR
∠B = ∠Q [proved above]
AB = PQ [Given]
BC = QR [Given]
∴ ∆ABC ≅ ∆PQR [By SAS congruency]
EX 7.3 QUESTION 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:
In ΔECB and ΔFBC,
∠BFC = ∠CBE [Each 90°]
BC = CB [Common]
FC = BE [Given]
∆ECB ≅ ∆FBC [ RHS congruency]
∠FBC = ∠ECB [C.P.C.T.]
⇒ AB = AC [Sides opposite to equal angles of a ∆ are equal]
∴ ABC is an isosceles triangle.
EX 7.3 QUESTION 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that B = C.
Solution:
In ∆ABP and ∆ACP,
∠APB = ∠APC [Each 90°]
AB = AC [Given]
AP = AP [Common]
∴ ∆ABP ≅ ∆ACP [By RHS congruency]
So, ∠B = ∠C [By C.P.C.T.]
