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Angles



Question (1): In two right triangles, an acute angle and a side of one are equal to the the angle and corresponding side of the other. Prove that the triangles are congruent.
Answer: Given: In triangles ABC and DEF, BC = EF To prove: Proof: In triangles ABC and DEF. BC = EF ( given)

Question (2): AB is a line segment AX and BY are two equal line segments drawn on opposite sides of line AB such that AX\|\|BY. If line segment AB and XY intersect each other at the point P, then prove that ii) a line segments AB and XY bisect each other at P.
Answer: Given: AB is a line segment AX\|\|BY AB and XY intersect at P To prove: ii) AP=PB and PY=PX Proof: Compare triangles APX and BPY, (given) AP = PB ( corresponding parts of congruent triangles) PY = PX

Question (3):
Answer: Given: AC=BC To prove: Proof: In triangles DBC and EAC, BC=AC (given) (given) From the figure, From the figure, From (i) and (ii),

Question (4): In the adjoining figure, AB ^ BD, DE ^ BD, BC = CD and
Answer: Given: BC=CD To prove: Proof: Compare triangles ABC and ECD, BC=CD ( given) (given) (given)

Question (5): In the adjoining figure, AB\|\|CD. BC and AD are transversals intersecting at O such that OA=OD. Show that
Answer: Given: AB\|\|CD. AD and BC intersect at O. OA = OD. To prove: Proof: In triangles AOB and COD, OA=OD (given) ( alternate angles) ( vertically opposite angles)

Question (6): In D PQR, P=40^o, Q=65^o and R=75^o. Name the greatest side and the least side.
Answer: (theorem on inequalities) (theorem on inequalities)

Question (7): In D XYZ, XY=6.5cm, YZ=5.5cm and XZ=7.5cm. Name the greatest and smallest angle.
Answer: (theorem on inequalities) (theorem on inequalities)

Question (8): In the adjoining figure PQR is an equilateral triangle. N is any point on PQ. Show that (i) NR>PN (ii) NR>NQ.
Answer: Given: In D PQR, PQ=PR=QR To prove: (i) NR>PN (ii) NR>NQ Proof: In D PQR, PQ=PR=QR ( given) ( Equilateral triangle is equiangular) In D PNR, ( is only a part of PRQ) ( given) NR>PN (theorem on inequalities) In DNQR, NR>NQ (theorem on inequalities)

Question (9): In a quadrilateral ABCD, AB=AD and BC>CD. Prove that . (Hint: Join BD)
Answer: Given: To prove: Construction: Join BD. Proof: In DBC, BC>CD ( given) ...(i) (theorem on inequalities) In D ABD, AB = AD ( given) ...(ii) By adding (i) and (ii), we get (adding equal thing to both sides does not alter inequation)

Question (10): In the adjoining figure D ABC, D is the mid point of BC and AD>BD or DC. Prove that .
Answer: Given: In D ABC, D is the mid-point of BC. AD>BD. To prove: Proof: AD>BD ( given) .....(i) (theorem on inequalities) AD>DC ( given) .....(ii) (theorem on inequalities) By adding (i) and (ii), we get

Question (11): In the adjoining figure, PQ>PR. QS and RS are the bisectors
Answer: Given: PQ>PR To prove: SQ>SR Proof: PQ > PR ( given) ( angle opposite to greater side) SQ > SR ( side opposite to greater side)

Question (12): In the adjoining figure x>y . Show that LM>LN.
Answer: Given: x>y To prove: LM>LN Proof: 180-smaller angle > 180^o- bigger angle i.e., (180-y) > (180-x) LM>LN side opposite to greater angle. LM > LN side opposite to greater angle.

Question (13): In the adjoining figure, PQ=PR. Show that PS>PQ.
Answer: Given: PQ=PR To prove: PS>PQ Proof: PQ=PR ( given) In D PQR, (by theorem on exterior angle) PS > PQ (side opposite to greater angles)

Question (14): In the adjoining figure, PR>PQ and PS is the bisector of . Show that x>y.
Answer: Given: PR>PQ To prove: x>y Proof:

Question (15): In the adjoining figure, AP ^ l and PR>PQ. Show that AR>AQ.
Answer: Given: AP ^ l and PR>PQ To prove: AR>AQ Construction: Draw AS such that PS=PQ. Proof: In D ASR, PS=PQ ( by construction) And AP ^ l (given) ( RHS postulate) AQ=AS (CPCT) ( angles opposite to equal sides) (Proved) AR > AQ (side opposite to greater angle)