**Triangles**

**Question 1.∆ABC and ∆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) ∆ABD ≅ ∆ACD**

(ii) ∆ABP ≅ ∆ACP

(iii) AP bisects ∠A as well as ∠D

(iv) AP is the perpendicular bisector of BC.

(ii) ∆ABP ≅ ∆ACP

(iii) AP bisects ∠A as well as ∠D

(iv) AP is the perpendicular bisector of BC.

**Solution:****(i) In ∆ABD and ∆ACD, we have**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, we have**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) As, ∆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

Hence, AP is the perpendicular bisector of BC.

**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 right ∆ABD and ∆ACD, we have**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) Since, ∆ABD ≅ ∆ACD,**⇒ ∠BAD = ∠CAD [By C.P.C.T.]

So, AD bisects ∠A

**Question 3.Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that**

**(i) ∆ABC ≅ ∆PQR(ii) ∆ABM ≅ ∆PQN**

**Solution:**In ∆ABC, AM is the median.

∴BM = 12 BC ……(1)

In ∆PQR, PN is the median.

∴ QN = 12QR …(2)

And BC = QR [Given]

⇒ 12BC = 12QR

⇒ BM = QN …(3) [From (1) and (2)]

**(i) In ∆ABM and ∆PQN, we have**AB = PQ , [Given]

AM = PN [Given]

BM = QN [From (3)]

∴ ∆ABM ≅ ∆PQN [By SSS congruency]

**(ii) Since ∆ABM ≅ ∆PQN**⇒ ∠B = ∠Q …(4) [By C.P.C.T.]

Now, in ∆ABC and ∆PQR, we have

∠B = ∠Q [From (4)]

AB = PQ [Given]

BC = QR [Given]

∴ ∆ABC ≅ ∆PQR [By SAS congruency]

**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:**

Since BE ⊥ AC [Given]

∴ BEC is a right triangle such that ∠BEC = 90°

Similarly, ∠CFB = 90°

Now, in right ∆BEC and ∆CFB, we have

BE = CF [Given]

BC = CB [Common hypotenuse]

∠BEC = ∠CFB [Each 90°]

∴ ∆BEC ≅ ∆CFB [By RHS congruency]

So, ∠BCE = ∠CBF [By C.P.C.T.]

or ∠BCA = ∠CBA

Now, in ∆ABC, ∠BCA = ∠CBA

⇒ AB = AC [Sides opposite to equal angles of a ∆ are equal]

∴ ABC is an isosceles triangle.

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

**Solution:**

We have, AP ⊥ BC [Given]

∠APB = 90° and ∠APC = 90°

In ∆ABP and ∆ACP, we have

∠APB = ∠APC [Each 90°]

AB = AC [Given]

AP = AP [Common]

∴ ∆ABP ≅ ∆ACP [By RHS congruency]

So, ∠B = ∠C [By C.P.C.T.]