We are asked to show that the of the bases of an are the same. Let's extend the diagram to include the information given in the question.
We need to show that BI is perpendicular to TP and that I is the midpoint of TP. Let's prove these properties one at a time.
According to , each pair of base angles of an isosceles trapezoid are congruent.
∠ RTP&≅ ∠ APT
∠ TRA&≅ ∠ PAR
We also know that IB is perpendicular to RA, so angles ∠ IBR and ∠ IBA are both right angles. This means that quadrilaterals ITRB and IPAB have three pairs of congruent angles. Since the sum of the angle measures is 360 in both quadrilaterals, we can conclude that the fourth angles are also congruent.
∠ TIB≅∠ PIB
Since T, I, and P are , these angles form a . Then, since these are congruent angles, Theorem 2-5 guarantees that they are both right angles. Therefore, we proved that BI is perpendicular to TP.
I is the midpoint of TP
Let's connect point I with vertices R and A and focus on triangles △ RIB and △ AIB.
Let's summarize what we know about these triangles.
| Congruence |
Justification
|
| BR≅BA |
BI bisects RA.
|
| ∠ RBI≅∠ ABI |
BI is perpendicular to RA.
|
Since BI is a common side, triangles △ RIB and △ AIB have two pairs of congruent sides and a congruent included angle. According to the , these two triangles are congruent. This also means that the third pair of sides and corresponding angle pairs are also congruent.
RI&≅AI
∠ BRI&≅∠ BAI
∠ BIR&≅∠ BIA
Let's mark the first two of these congruences on the diagram and shift our focus on triangles △ RIT and △ AIP.
Since TRAP is an isosceles trapezoid, according to Theorem 6-19 its base angles are congruent.
∠ BRT&≅∠ BAP
Let's also mark this congruence on the diagram.
Let's summarize what we know about triangles △ RIT and △ AIP.
| Congruence |
Justification
|
| RT≅AP |
Given (TRAP is an isosceles trapezoid)
|
| RI≅AI |
Proven above
|
| ∠ IRT≅∠ IAP |
Difference of congruent angle pairs
|
We see that triangles △ RIT and △ AIP have two pairs of congruent sides and a congruent included angle, so
according to the Side-Angle-Side (SAS) Congruence Postulate, these two triangles are congruent. This also means that the third pair of sides are also congruent.
TI≅PI
This means that I is the midpoint of TP. Together with the previous part, we now completed proving that BI is a perpendicular bisector of TP. We can summarize the process above in a .
Completed Proof
2
&Given:&& TRAPis a trapezoid
& &&TR≅PA
& &&BIperpendicular toRA
& &&BR≅BA
& &&T,I,andPare collinear
& &&R,B,andAare collinear
&Prove:&& BIis the perpendicular bisector ofTP
Proof:
