Higher Ability Questions of Comp. Mathematics for SEE Students.

For Q.N.24 Geometry 

1. In the given figure, AP丄BC,BR丄AC and CQ丄 AB. Prove that ∠OPQ = ∠OPR (दिइएको चित्रमा AP丄BC,BR丄AC र CQ丄 AB छन् भने, प्रमाणित गर्नुहोस:  ∠OPQ = ∠OPR)
Solution:
Given: In the given figure: AP丄BC,BR丄AC and CQ丄 AB
To prove: ∡OPQ = ∡OPR
                                                            Proof:

Statements                                                                         Reasons

1. PORC and POQB is a cyclic quadrilateral    [being exterior of quadrilateral angle is equal to interior opposite angle ]
2. B,C,R and Q  are con-cyclic points [ being two equal angles (∠BQC=∠BRC=90) standing on same line segment BC ]  
3. ∠OPR=∠OCR [Inscribed angles standing on same arc OR]
4. ∠OPQ=∠OBQ [inscribed angles standing on same arc OQ]
5. ∠OBQ=∠OCR[Inscribed angle standing on same arc QR]
6. ∠OPQ=∠OPR [ from statements from statements 3,4 and 5 ] Proved.

2.In the given figure, D, E and F are the midpoints of AB, AC and BC respectively and AG 丄BC then prove that DEFG is cyclic quadrilateral. (दिइएको चित्रमा Δ ABC को भुजा AB, AC र BC का मध्यबिन्दु हरु क्रमसः D, E र F तथा AG 丄BC भए DEFG चक्रिय चतुर्भुज हो भनि प्रमाणित गर्नुहोस)
Solution:
Given:In the given figure, D, E and F are the midpoints of AB, AC and BC respectively and AG 丄BC

To prove: DEFG is cyclic quadrilateral.

                                                            Proof:
   Statements                                                                         Reasons

1. ∠ABG=90    [ BeingAG 丄 BC]
2. AD=BD=DG [Midpoint of hypotenuse of right angled triangle is equidistant from vertices ]  
3. DE॥BC and BA॥EF [The line joining the midpoint of two sides of triangle is parallel to third side]
4. BFED is a parallelogram [Being opposite sides parallel]
5. ∠DBG= ∠DGB [Being BD=DG, base angles of isosceles triangle]
6. ∠DBG= ∠DEF [ Opposite angles of parallelogram]
7. ∠DGB= ∠DEF [From statements 5 and 6]
8. DEFG is cyclic quadrilateral [ Being exterior angle of quadrilateral is equal to interior opposite angle]       Proved

3. In the given figure, O is the center of circle where XO॥SQ, prove that PX=XS. (दिइएको चित्रमा o वृतको केन्द्र र  XO॥SQ भए प्रमाणित गर्नुहोस PX=XS)

Solution:

Given: O is the center of circle and XO॥SQ.

To prove:PX=XS

Construction: Join S and O

                                          Proof:
         Statements                                              Reasons

1. OX=OQ    [ Being radii of circle ]
2. ∠OQS=∠OSQ [Being OS=OQ]  
3. ∠OQS=∠POX [Corresponding angles as XO॥SQ]
4. ∠XOS=∠OST[Alternate angles as XO॥SQ]
5. ∠POX= ∠XOS [From statement 2,3 and 4]
6. In △PXO and △OXS

          i) PO=OS (S) [ radii of circle]

          ii) ∠POX= ∠XOS (A) [ From statement 5]

          iii)OX=OX (S) [ Common side]

     7. △PXO ≌ △OXS [ By S.A.S axiom ]

     8. PX=XS [ Corresponding sides of congruent triangles            are equal.]

       Proved.

4. WXYZ is a cyclic quadrilateral, if the bisectors of ∠XWZ and ∠XYZ meets the circle at points P and Q respectively, then prove that PQ is a diameter of circle.[WXYZ एउटा चक्रिय चतुर्भुज हो | यदि ∠XWZ र ∠XYZ का अर्धकहरुले वृत्तलाई क्रमशः P र Q मा भेट्छन भने PQ  वृत्तको ब्यास हो भनी प्रमाणित गर्नुहोस |

Solution:
Given: 
WXYZ is a cyclic quadrilateral, bisectors of ∠XWZ and ∠XYZ meets the circle at points P and Q respectively.
To Prove:
PQ is a diameter of circle.
                                              Proof:

         Statements                                              Reasons

1. ∠ZWP=∠PWX    [ PW bisects ∠XWZ]
2. Arc(PZ)= arc(PX) [Opposite arcs of equal angles are equal]  
3. ∠ZYQ=∠QYX [QY bisects ∠XYZ]
4. Arc(ZQ) = Arc(XQ)[Same as in statement 2.]
5. Arc(PZ)+Arc(ZQ) = Arc(PX)+Arc(XQ) [Adding statement 2 and 4 ]
6. Arc(PZQ) = Arc(PXQ) [Whole part axiom]
7. PQ is a diameter [From statement 6, being arc on both sides of PQ equals]

    Proved.