**1. **Using basic proportionality theorem, prove that a line drawn through the mid point of one side of a triangle parallel to another side bisects the third side.

**Proof: **Let D is the mid point of AB and DE is parallel to BC.

To prove: E is the mid point of AC.

**Proof: **Since DE is parallel to BC, therefore by basic proportionality theorem [latex]\frac{AB}{AD}=\frac{AC}{AE}[/latex]

[latex]\frac{AB}{AD}-1=\frac{AC}{AE}-1[/latex]

[latex]\frac{BD}{AD}=\frac{CE}{AE}

[latex]\frac{BD}{BD}=\frac{CE}{AE} ( BD = AD )

CE = AE

E is the mid-point of AC.

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

**Proof: **In triangle ABD and triangle ADC

BD = CD (D is the mid point of BC)

AD = AD ( common side )

Angle ADB = Angle ADC (90 degree)

Hence triangle ADB is congruent to ADC ( by SAS property)

hence AB = AC

**3. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA ^{2} = CB.CD.**

**Solution:** In ΔBAC and ΔADC;

∠BAC = ∠ADC (given)

∠ACB = ∠DCA (Common angle)

Hence ΔBAC ~ ΔADC

Hence [latex]\frac{CA}{CB}=\frac{CD}{CA}[/latex]

[latex]\Rightarrow[/latex]CA x CA = CB x CD

[latex]\Rightarrow[/latex]CA^{2} = CB x CD proved

**4. **ABC is a right angled triangle, right angle at B and BD is perpendicular to AC. If AB = 6, BC = 8 and AC = 10, then BD = ?

**Solution: **

[latex]\angle[/latex]C = [latex]\angle[/latex]C

[latex]\angle[/latex]B = [latex]\angle[/latex]D

Hence [latex]\triangle[/latex] ABC is similar to BDC

Therefore, [latex]\frac{AB}{BD}=\frac{AC}{CD}[/latex]

[latex]\frac{6}{BD}=\frac{10}{6}[/latex]

So BD = 3.6

**5. **ABC is a triangle in which DE is parallel to BC. If AD = 4, BD = 3, AE = 12, find the value of AC.

**Solution: **Since DE is parallel to BC, therefore [latex]\frac{AD}{BD}=\frac{AE}{CE}[/latex]

[latex]\frac{4}{3}=\frac{12}{x}[/latex]

x = 9

so AC = 12 + 9 = 21