PQRS is a llgram where X and Y are the mid points of the side PQ and RS respectively. W and Z are points of intersection of SX, PY, XR and YQ respectively. Show that: ar( YWZ) = ar(XWZ)
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Since PQRS is a ∥gm, then
PQ∥SR and PS∥QR.
Since SX bisects ∠PSR, then
∠PSX = ∠XSR .....(1)
Since SR∥PQ, and SX is a transversal, then
∠XSR = ∠PXS [Alternate interior angles] .....(2)
From (1) and (2), we get
∠PSX = ∠PXS
In ∆PSX,
∠PSX = ∠PXS
⇒PX = PS (Sides opposite to equal angles are equal) ....(3)
Now, X is the mid point of PQ, so
PX = XQ ........(4)
Now, QR = PS (Opposite sides of ∥gm are equal) ......(5)
From (3), (4) and (5), we get
XQ = QR
In ∆XQR,
XQ = QR
⇒∠XRQ = ∠QXR [Angles opposite to equal sides are equal] .........(6)
Since SR∥PQ and RX is a transversal, then∠XRS = ∠QXR (Alternate interior angles) .......(7)Now, from (6) and (7), we get∠XRS = ∠XRQ⇒RX bisects ∠SRQ.In ∆SXR,∠XSR + ∠XRS + ∠SXR = 180° [Angle sum property]⇒12∠XSR + 12∠XRS + 12∠SXR = 90°⇒12∠SXR = 90° − 12(∠XSR + ∠XRS) ........(8)We know that adjacent angles of ∥gm are supplementarySo, ∠S + ∠R = 180°⇒12(∠S + ∠R) = 90°⇒∠XSR + ∠XRS = 90°Now, from (8), 12∠SXR = 90° − 12[90°] = 45°⇒∠SXR = 90°
PQ∥SR and PS∥QR.
Since SX bisects ∠PSR, then
∠PSX = ∠XSR .....(1)
Since SR∥PQ, and SX is a transversal, then
∠XSR = ∠PXS [Alternate interior angles] .....(2)
From (1) and (2), we get
∠PSX = ∠PXS
In ∆PSX,
∠PSX = ∠PXS
⇒PX = PS (Sides opposite to equal angles are equal) ....(3)
Now, X is the mid point of PQ, so
PX = XQ ........(4)
Now, QR = PS (Opposite sides of ∥gm are equal) ......(5)
From (3), (4) and (5), we get
XQ = QR
In ∆XQR,
XQ = QR
⇒∠XRQ = ∠QXR [Angles opposite to equal sides are equal] .........(6)
Since SR∥PQ and RX is a transversal, then∠XRS = ∠QXR (Alternate interior angles) .......(7)Now, from (6) and (7), we get∠XRS = ∠XRQ⇒RX bisects ∠SRQ.In ∆SXR,∠XSR + ∠XRS + ∠SXR = 180° [Angle sum property]⇒12∠XSR + 12∠XRS + 12∠SXR = 90°⇒12∠SXR = 90° − 12(∠XSR + ∠XRS) ........(8)We know that adjacent angles of ∥gm are supplementarySo, ∠S + ∠R = 180°⇒12(∠S + ∠R) = 90°⇒∠XSR + ∠XRS = 90°Now, from (8), 12∠SXR = 90° − 12[90°] = 45°⇒∠SXR = 90°
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