Please solve this, with full explaination.
Answers
Given:
QX bisects ∠Q.
i.e, ∠TQX = ∠SQX
RX bisects ∠R.
i.e, ∠SRX = ∠PRX
To prove:
(i) ΔXTQ ≅ ΔXSQ
(ii) PX bisects ∠P
Construction:
Join P to X.
Draw XA perpendicular to PR.
Solution:
We've been given that QX bisects ∠Q, this means that QX divides ∠Q into two equal parts/angles. Therefore: ∠TQX = ∠SQX
Similarly, RX bisects ∠R, this means that RX divides ∠R into two equal parts/angles. Therefore: ∠SRX = ∠PRX
(i) Now, in ΔXTQ and ΔXSQ:
XQ = XQ (Common side to both triangles)
∠XTQ = ∠XSQ = 90° (Both angles are equal to 90°; Given)
∠TQX = ∠SQX (Given/Proved)
∴ By using AAS (Angle-Angle-Side) congruency criterion, we can say that:
⇒ ΔXTQ ≅ ΔXSQ
By using CPCT (Corresponding Parts of Congruent Triangles)
⇒ XS = XT → Relation(1)
(ii) Now, we are asked to show that PX bisects ∠P.
In simple words, we're asked to show that ∠TPX = ∠RPX.
For this, Join P to X, forming a line PX.
Draw XA ⊥ to PR.
In ΔXAR & ΔXSR
XR = XR (Common side)
∠XSR = ∠XAR = 90° (Construction)
∠SRX = ∠ARX (RX bisects ∠R)
∴ By using AAS (Angle-Angle-Side) congruency criterion, we can say that:
⇒ ΔXAR ≅ ΔXSR
By using CPCT (Corresponding Parts of Congruent Triangles)
⇒ XS = XA → Relation(2)
From Relation (1) & Relation (2):
XS = XA & XS = XT
Therefore, XA = XT → Relation(3)
(Things equal to the same thing are equal to one another)
Now, In ΔXAP & ΔXTP
XP = XP (Common side)
∠XTP = ∠XAP = 90° (Given & Construction)
XA = XT (From Relation 3)
∴ By using RHS (Right Angle-Hypotenuse-Side) congruency criterion, we can say that:
⇒ ΔXAR ≅ ΔXTP
By using CPCT (Corresponding Parts of Congruent Triangles)
⇒ ∠TPX = ∠APX.
Since these angles are equal, we can say that XP divides the ∠P into two equal parts, making XP the angle bisector of ∠P.
Therefore, PX bisects ∠P.
Hence proved.
(Refer to the attachment for markings, the lines in red are the ones constructed by us)
Answer:
check this years question paper
