Thread: If a quadrilateral is circumscribable, then the sum of the lengths of two opposite si

Good day,.,.can anyone please help me with the proof of this???thnx

If a quadrilateral is circumscribable, then the sum of the lengths of two opposite sides equals the sum of the lengths of the sides of the two remaining sides.

Originally Posted by aldrincabrera

If a quadrilateral is circumscribable, then the sum of the lengths of two opposite sides equals the sum of the lengths of the sides of the two remaining sides.[/I]

Do you know the external tangent theorem: tangents from an exterior point to a circle?

,.,.that's the thing that is making me confuse sir,.,coz i can't somehow imagine the diagram,.isn't the sum of two opposite sides longer than the sum of the other two???im confused,.,.can u help me sir??thnx

Originally Posted by aldrincabrera

,.,.that's the thing that is making me confuse sir,.,coz i can't somehow imagine the diagram,.isn't the sum of two opposite sides longer than the sum of the other two???im confused,.,.can u help me sir??thnx

Go to this page.
Look at the first diagram. You want to prove that

,.,thnx sir,,.does this mean (always) that side a is congruent to side d??and b with c???

Originally Posted by aldrincabrera

,.,thnx sir,,.does this mean (always) that side a is congruent to side d??and b with c???

In that diagram side a is the sum of two different line segments each tangent from external points. One those is also contained in b and the other is in d.

,.,.thanks so much sir,.,now i have a clear perception for the proof im going to make,.,.thnx