Q)15 .....please prove....help....Mates
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AP, AQ, BP and BQ are the angle bisectors of ∠ mAB, ∠ BAn, ∠ ABx and ∠ zBA.
Here,
mn || xz and T is transversal.
∴ ∠ mAB = ∠ zBA (alternate angles)
⇒ 1/2 ∠ mAB = 1/2 ∠ zBA
⇒ ∠alpha = ∠ beta (∠ alpha = 1/2 ∠ mAB and ∠ beta = 1/2 ∠ zBA)
Therefore, PA || BQ ….....(1)
Also ∠ nAB = ∠ xBA (alternate angles)
⇒ 1/2 ∠nAB = 1/2 ∠ xBA
⇒ ∠ gamma = ∠ delta (∠ gamma = 1/2 ∠ nAB and ∠ delta = 1/2 ∠ xBA)
Therefore, PB || QA …....(2)
From (1) and (2), we get
PAQB is a parallelogram ....(3)
Now,
∠ xBz= 180°
⇒ 1/2 ∠ xBz = 180/2 = 90°
⇒ 1/2 (∠xBA + ∠ zBA) = 90°
⇒ 1/2 ∠ xBA + 1/2 ∠ zBA = 90°
⇒ ∠delta + ∠ beta = 90°
⇒ ∠ PBQ = 90° ...(4)
So, using (3) and (4), we conclude that PBQA is a rectangle.
QED