Question

In a quadrilateral PQRS, the bisectors of angle R and angle S
meet at O point T. Show that angle P + angle Q = 2 angle RTS.
​

Answer 1

Step-by-step explanation:

We know that sum of interior angles of quadrilateral is 360o

∠P+∠Q+∠R+∠S=360o

∠P+∠Q+100+50=360

∠P+∠Q=210o ……………..(1)

OP and OQ are bisectors of angles P and Q,

∠P=2∠OPQ and ∠Q=2∠OQP

So from (i), we have

2∠OPQ+2∠OQP=210o

2(∠OPQ+∠OQP)=210o

∠OPQ+∠OQP=105o …………….(2)

Consider triangle POQ,

∠POQ+∠OPQ+∠OQP=180o

∠POQ+105o=180o [using equation (3)]

∠POQ=180o−105o

∠POQ=75o.

Answer 2

★ Given :-

• PQRS is a quadrilateral.
• The bisectors of ∠R and ∠S meet at point T

★ To Prove :-

• ∠P + ∠Q = 2 ∠RTS

★ Diagram :-

$\setlength{\unitlength}{1cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(7,1)\qbezier(1,1)(1,1)(0,4)\qbezier(7,1)(7,1)(6,4)\qbezier(6,4)(6,4)(0,4)\qbezier(6,4)(6,4)(3.7,2.5)\qbezier(0,4)(0,4)(3.7,2.5)\put(1,0.5){\large\sf P}\put(7,0.5){\large\sf Q}\put(5.8,4.3){\large\sf R}\put(-0.3,4.3){\large\sf S}\put(3.6,2){\large\sf T}\end{picture}$

$\boxed {\star{ \underline{ \red{ \sf \: Solution :-}}}}$

$\sf \: In \: \triangle \: STR, we \: have,$

$\sf\angle STR \: + \angle \: TRS \: + \angle \: TSR = 180 \degree$

$\implies \: \sf\angle STR \: = 180 \degree \: - \sf ( \angle \: TRS \: + \angle \: TSR)$

$\implies \: \sf\angle STR \: = 180 \degree \: - \sf ( \dfrac{1}{2} \angle \: R \: + \dfrac{1}{2} \angle \: S)$

$\implies \sf\angle STR \: = 180 \degree - \dfrac{1}{2} ( \angle \: R \: + \angle \: S)$

$\implies \sf\angle STR = 180 \degree - \dfrac{1}{2} [360 \degree - ( \angle \: P \: + \angle \: Q)]$

$\implies \sf\angle STR \: = 180 \degree \: - 180 \degree + \dfrac{1}{2} ( \angle \: P \: + \angle \: Q)$

$\implies \sf\angle STR \: = \dfrac{1}{2} ( \angle \: P \: + \angle \: Q)$

$\implies \sf2\angle STR \: = ( \angle \: P \: + \angle \: Q)$

$\sf \: \therefore \: ( \angle \: P \: + \angle \: Q) = \sf2\angle STR \: (proved)$