Question

13. In Figure 13, PQRS is a quadrilateral
in which PQ|| SR and PS||QR.
Prove that angle SPQ = angle SRQ.​

Answer 1

Answer:

$\red{ \underline{ \huge \bold{Given: }}}$

• PQRS is a quadrilateral
• in which PQ|| SR and PS||QR.

$\red{ \underline{ \huge \bold{To \: prove: }}}$

• Angle SPQ = angle SRQ.

$\red{ \underline{ \huge \bold{Solution: }}}$

$\sf \: Given \: that \: PQ||SR \\ </p><p> \sf \:and \: PS \: intersect \: both \: parallel \: lines \\ </p><p> \sf \: = > ∠SPQ+∠PSR=180° \: \: \: \: \pink{ Eq 1 }\\ </p><p> \\ </p><p> \sf \:PS||QR \: is \: also \: given \\ </p><p> \sf \:and \: SR \: intersect \: both \: lines \\ </p><p> \sf \: = > ∠PSR+∠SPQ =180° \: \: \: \: \: \pink{Eq 2}$

$\sf \: Equating \: Eq 1 \: and \: Eq 2 \\ </p><p></p><p> \sf \:∠SPQ+∠PSR=∠PSR+∠SRQ \\ </p><p> \sf \:Cancelling \: ∠PSR \: from \: both \: sides \\ \\ \sf∠SPQ+ \cancel{∠PSR}=\cancel{∠PSR}+∠SRQ \\ </p><p> \sf \:∠SPQ=∠SRQ \\ </p><p> \pink{\sf \:QED ☆</p><p>}\\ </p><p> \pink{ ☆</p><p>\sf \:Hence \: proved}$

Answer 2

Answer:

Answer:

\red{ \underline{ \huge \bold{Given: }}}

Given:

PQRS is a quadrilateral

in which PQ|| SR and PS||QR.

\red{ \underline{ \huge \bold{To \: prove: }}}

Toprove:

Angle SPQ = angle SRQ.

\red{ \underline{ \huge \bold{Solution: }}}

Solution:

\begin{gathered}\sf \: Given \: that \: PQ||SR \\ \sf \:and \: PS \: intersect \: both \: parallel \: lines \\ \sf \: = > ∠SPQ+∠PSR=180° \: \: \: \: \pink{ Eq 1 }\\ \\ \sf \:PS||QR \: is \: also \: given \\ \sf \:and \: SR \: intersect \: both \: lines \\ \sf \: = > ∠PSR+∠SPQ =180° \: \: \: \: \: \pink{Eq 2}\end{gathered}

GiventhatPQ∣∣SR

andPSintersectbothparallellines

=>∠SPQ+∠PSR=180°Eq1

PS∣∣QRisalsogiven

andSRintersectbothlines

=>∠PSR+∠SPQ=180°Eq2

\begin{gathered}\sf \: Equating \: Eq 1 \: and \: Eq 2 \\ \sf \:∠SPQ+∠PSR=∠PSR+∠SRQ \\ \sf \:Cancelling \: ∠PSR \: from \: both \: sides \\ \\ \sf∠SPQ+ \cancel{∠PSR}=\cancel{∠PSR}+∠SRQ \\ \sf \:∠SPQ=∠SRQ \\ \pink{\sf \:QED ☆ }\\ \pink{ ☆ \sf \:Hence \: proved}\end{gathered}

EquatingEq1andEq2

∠SPQ+∠PSR=∠PSR+∠SRQ

Cancelling∠PSRfrombothsides

∠SPQ+

∠PSR

=

∠PSR

+∠SRQ

∠SPQ=∠SRQ

QED☆

☆Henceproved