# Congurency In Triangles

## Exercise 7.2

Question 1: In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠ C intersect each other at O. Join A to O. Show that: (a) OB = OC (b) AO bisects ∠ A.

∠OBC = ∠ OAC (They are halves of angless B and C)
So, OB = OC (Sides opposite to equal angles)

In ΔAOB and ΔAOC
AB = AC (given)
OB = OC (proven earlier)
∠ABO = ∠ACO (they are halves of angles B and C)
So, ΔAOB ≅ ΔAOC(SAS rule)
So, ∠BAO=∠CAO
It means that AO bisects ∠A

Question 2: In Δ ABC, AD is the perpendicular bisector of BC. Show that Δ ABC is an isosceles triangle in which AB = AC.

Answer: In ΔABD and ΔACD
AD = AD (common side)
BD = CD (given)
∠ADB=∠ADC (right angle)
So, ΔABD≅ ΔACD
So, AB = AC
This proves that ΔABC is isosceles triangle.

Question 3: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.

Answer: In ΔABE and ΔACF
AB = AC (given)
∠BAE=∠CAF (common to both triangles)
∠CFA=∠BEQ (right angles)
So, ΔABE≅ ΔACF (ASA rule)
So, BE=CF

Question 4: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that

1. ΔABE≅ ΔACF
2. AB = AC, i.e., ABC is an isosceles triangle.

Answer: This can be solved like previous question.

Question 5: ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.

Answer: ∠ABC=∠ACB
∠DBC=∠DCB
So, ∠ABC+∠DBC=∠ACB+∠DCB
Or, ∠ABD=∠ACD

Question 6: ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.

AD=AB
AC=AC
∠ACB=∠ABC
∠ACD=∠ACD
In ΔABC, ∠ACB+∠ABC+∠CAB=180°
Or, ∠CAB=180°-2∠ACB ----- (1)

∠DAC=180°-2∠ACD ------ (2)
As BD is a straight line, so ∠CAB+∠DAC=180°
So, adding equations (1) and (2) we get
180°=360°-2∠ACB-2∠ACD
Or, 180°=360°-2(∠ACB+∠ACD)
Or, 2(∠ACB+∠ACD)=180°
Or, ∠ACB+∠ACD=∠BCD=90°

Question 7: ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Answer: If AB = AC then angles opposite to these sides will be equal. As you know the sum of all angles of a triangle is equal to 180°.
So, ∠A+∠B+∠C=190°
Or, 90°+∠B+∠C=180°
Or, ∠B+∠C=180°-90°=90°
Or, ∠B=∠C=45°

Question 8: Show that the angles of an equilateral triangle are 60° each.

Answer: As angles opposite to equal sides of a triangle are always equal. So, in case of equilateral triangle all angles will be equal. So they will measure one third of 180°, which is equal to 60°