# 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.

**Answer:** In ΔOBC

`∠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

- `ΔABE≅ ΔACF`
- 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.

**Answer:** In ΔADC and ΔABC

`AD=AB`

`AC=AC`

`∠ACB=∠ABC`

`∠ACD=∠ACD`

In ΔABC, `∠ACB+∠ABC+∠CAB=180°`

Or, `∠CAB=180°-2∠ACB` ----- (1)

Similarly, in ΔADC

`∠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°