Question

PLEASE HELP ASAP (6 POINTS EACH)

In △ABC, point D is on side AB so that CA=CD. Prove that BC>AC.

Answer 1

Given: In
`\triangle ABC`, point D is on the side AB so that CA=CD.

Isosceles Triangle states:

* A triangle with two equal sides.

*The angles which are opposite to the equal sides are also equal.

Since, CA=CD

By definition of isosceles triangle,

`\triangle CAD` is an isosceles triangle,

All the three angles within an isosceles triangle are acute. Acute angles are always less than 90 degrees.

therefore,
`\angle CAD=\angle CDA`
`<90^(\circ)` ......[1]

Now, Consider
`\triangle CBD`,

Exterior Angle Property state that an exterior angle of a triangle is equal to the sum of the opposite interior angles.

Therefore, by exterior angle property in
`\triangle CBD` we have ;

`\angle CBD + \angle BCD = \angle CDA`

So,
`\angle CBD< \angle CDA`

From [1] we have:

`\angle CBD < \angle CDA=\angle CAD <90^(\circ)`

Consider
`\triangle CAB`,

Sine law states that an equation relating the lengths of the sides of a triangle to the sines of its angles.

Now, using Sine law in
`\triangle CAB`,

`(\sin (\angle CAD))/(BC) =(\sin (\angle CBD))/(AC)` ......[2]

Since,

`\angle CBD < \angle CDA< 90^(\circ)`

so,
`\sin (\angle CBD)<\sin (\angle CAD)` ......[3]

From [2] and [3] we have;

`AC<BC` or we can write it as
`BC>AC` hence Proved!