Line XD equals line XE and is also equal to line XF that proved by using AAS postulate of congruence ⇒ B
Step-by-step explanation:
Let us revise the cases of congruence
- SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ
- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
- AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles and one side in the 2nd Δ
- HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ
In Δ ABC
∵ Ray AL bisects ∠A ⇒ (divides it into two equal angles)
∴ m∠DAX = m∠EAX
∵ Ray BM bisects ∠B ⇒ (divides it into two equal angles)
∴ m∠EBX = m∠FBX
∵ XD ⊥ AC
∴ m∠XDA = 90°
∵ XE ⊥ AB
∴ m∠XEA = 90°
∵ XE ⊥ BC
∴ m∠XFB = 90°
Now lets prove that Δ ADX and ΔAEX are congruent
In Δs ADX and AEX
∵ m∠ADX = m∠AEX ⇒ (their measures are 90°)
∵ m∠DAX = m∠EAX ⇒ proved
∵ AX is a common side in both triangles
- By using the AAS postulate of congruence
∴ Δ ADX ≅ Δ AEX
∴ XD = XE
Let us do the same with Δ BEX and Δ BFX
In Δs BEX and BFX
∵ m∠BEX = m∠BFX ⇒ (their measures are 90°)
∵ m∠EBX = m∠FBX ⇒ proved
∵ BX is a common side in both triangles
- By using the AAS postulate of congruence
∴ Δ BEX ≅ Δ BFX
∴ XE = XF
∵ XE = XD
∵ XE = XF
- If one side is equal two other sides then the two other sides are
equal, that means the three sides are equal
∴ XD = XF
∴ XD = XE = XF
Line XD equals line XE and is also equal to line XF that proved by using AAS postulate of congruence
Learn more:
You can learn more about the congruence in brainly.com/question/6108628
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