Mathematics
In the given figure, equal sides BA and CA of △ABC are produced to Q and P respectively such that AP = AQ. Prove that PB = QC.
Answer
In △ ABC,
⇒ BA = CA [Given]
⇒ ∠ABC = ∠ACB [Angles opposite to equal sides are equal]
From figure,
⇒ ∠PAB + ∠CAB = 180° [Linear pair of angles]
⇒ ∠PAB = 180° - ∠CAB ……(1)
⇒ ∠QAC + ∠CAB = 180° [Linear pair of angles]
⇒ ∠QAC = 180° - ∠CAB ……(2)
From eq.(1) and (2), we have:
⇒ ∠PAB = ∠QAC
In △PAB and △QAC,
⇒ ∠PAB = ∠QAC [Proved above]
⇒ AP = AQ [Given]
⇒ BA = CA [Given]
∴ △PAB ≅ △QAC (By S.A.S axiom)
∴ PB = QC [Corresponding angles of congruent triangles are equal]
Hence, proved that PB = QC.
Related Questions
Which of the following pairs of triangles are congruent ?
(a) △ABC and △DEF in which : BC = EF, AC = DF and ∠C = ∠F.
(b) △ABC and △PQR in which : AB = PQ, BC = QR and ∠C = ∠R.
(c) △ABC and △LMN in which : ∠A = ∠L = 90°, AB = LM, ∠C = 40° and ∠M = 50°
(d) △ABC and △DEF in which : ∠B = ∠E = 90°, AC = DF
In the given figure, P is a point in the interior of ∠ABC. If PL ⊥ BA and PM ⊥ BC such that PL = PM, prove that BP is the bisector of ∠ABC.
In the given figure, median AD of △ABC is produced. If BL and CM are perpendiculars drawn on AD and AD produced, prove that BL = CM.
In the given figure, M is the mid-point of AB and CD. Prove that CA = BD and CA || BD.
