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Exercise 7.3

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 figure). If AD is extended to intersect BC at P, show that

(i)  ΔABD  ≅  ΔACD

Sol. –
Consider triangles ABD and ACD,

We have AB = AC                                                [Given]

BD = CD                                                                [Given]

So, ΔABD ≅ ΔACD                                              [SSS rule]

(ii) ΔABP ≅  ΔACP

Sol. –
Consider triangles ABP and ACP,

We have AB = AC                                                          [Given]

AP = PA                                                                          [Common]

and  ∠BAP = ∠CAP                                                      [From (i)]

∴             ΔABP  ≅ ΔACP                                             [SAS rule]

⇒          BP = PC,   ∠BPA = ∠CPA

(iii) AP bisects ∠A as well as ∠D.

Sol. –

∠BAP = ∠CAP                                [From result (ii)]

⇒     AP bisects ∠A.

⇒            ∠BDP = ∠CDP                                       [Exterior angle property]

So, DP bisects ∠D.

Hence, AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC.

Sol.  –
Also, ∠BPA + ∠CPA = 180°                        [Linear pair]

⇒       2∠BPA = 180°

⇒      ∠BPA = 90°                                     [From result (ii)]

As BP = CP                                                [From result (ii)]

and       AP ⊥ BC                                       [Proved above]

⇒         AP is perpendicular bisector of BC.

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 figure). Show that:
(i) ΔABM ≅ ΔPQN
(ii) ΔABC  ≅ ΔPQR

Sol. (i) M and. N are mid-points of Be and QR respectively, as AM and PN are medians.

$BM = {1 \over 2}BC$ and $QN = {1 \over 2}QR$                        …..(i)

Also, BC = QR                                         [ Given ]

$\Rightarrow {1 \over 2}BC = {1 \over 2}QR \Rightarrow BM = QN$    …..(ii)   [From (i)]

Consider triangles ABM and PQN,

We have            AB = PQ                                         [ Given ]

AM=PN                                                                     [ Given ]

and BM = QN                                                          [ From (ii)]

∴   ΔABM ≅ ΔPQN                                                 [SSS rule ]
(ii) From result (i),

∠ABM = ∠PQN  ⇒ ∠ABC = ∠PQR                           ……(iii)

Consider triangles ABC and PQR,

We have AB = PQ                                                     [Given]

BC = QR                                                                      [Given]

and   ∠ABC = ∠PQR                                                [From (iii).]

∴       ΔABC  ≅ ΔPQR.                                              [SAS rule]

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

Sol.
Consider triangles BFC and CEB,

We have BE = GE                                                          [Given]

BC = CB                                                                           [Common]

and ∠BEC = ∠BEC                                                        [90° each]

∴     ΔBFC =  ΔCEB                                                        [RHS rule]

⇒        ∠FBC = ∠ECB                                                     [CPCT]

i.e.,     ∠ABC = ∠ACB

⇒     AC = AB

[Sides opposite to equal angles of a triangles are equal.]

⇒   ΔABC is an isosceles triangle.

5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Sol. Consider triangles APB and APC,

We have           AB = AC                      [ Given ]

AP = PA                                                 [ Common ]

∠APB = ∠APC                                      [90° each ]

∴       ΔABP   ≅   ΔACP                        [ RHS rule ]

⇒        ∠B = ∠C.                                     [ CPCT ]