Hi,
33)

Let ABC be a triangle having sides AB, BC and CA.

We know that the sum of any two sides of a triangle is greater than the third side.

∴ AB + BC > CA

⇒AB > CA – BC

This shows that difference of two sides is less than the third side of the triangle.

Similarly, we can  prove it for other sides as well.

Regards

• 1

In ΔABC,

AB = AC (Given)

⇒ ACB = ABC (Angles opposite to equal sides of a triangle are also equal)

In ΔACD,

⇒ ADC = ACD (Angles opposite to equal sides of a triangle are also equal)

In ΔBCD,

ABC + BCD + ADC = 180º (Angle sum property of a triangle)

⇒ ACB + ACB +ACD + ACD = 180º

⇒ 2(ACB + ACD) = 180º

⇒ 2(BCD) = 180º

⇒ BCD = 90º

• 5

thanks

• 4

In ΔABC,

AB = AC (Given)

⇒ ACB = ABC (Angles opposite to equal sides of a triangle are also equal)

In ΔACD,

⇒ ADC = ACD (Angles opposite to equal sides of a triangle are also equal)

In ΔBCD,

ABC + BCD + ADC = 180º (Angle sum property of a triangle)

⇒ ACB + ACB +ACD + ACD = 180º

⇒ 2(ACB + ACD) = 180º

⇒ 2(BCD) = 180º

⇒ BCD = 90º

• 3
AB = AC (Given)

⇒ ACB = ABC (Angles opposite to equal sides of a triangle are also equal)

In ΔACD,

⇒ ADC = ACD (Angles opposite to equal sides of a triangle are also equal)

In ΔBCD,

ABC + BCD + ADC = 180º (Angle sum property of a triangle)

⇒ ACB + ACB +ACD + ACD = 180º

⇒ 2(ACB + ACD) = 180º

⇒ 2(BCD) = 180º

⇒ BCD = 90º
• 1
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